problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
8.059. $2 \operatorname{tg}^{4} 3 x-3 \operatorname{tg}^{2} 3 x+1=0$.
8.059. $2 \tan^{4} 3 x-3 \tan^{2} 3 x+1=0$. | ## Solution.
Domain of definition: $\cos 3 x \neq 0$.
Solving this equation as a biquadratic equation in terms of $\operatorname{tg} 3 x$, we get: 1) $\left.\operatorname{tg} 3 x= \pm \frac{\sqrt{2}}{2}, x_{1,2}=\frac{ \pm \operatorname{arctg} \frac{\sqrt{2}}{2}}{3}+\frac{\pi k}{3}, k \in Z ; 2\right) \operatorname{t... | x_{1,2}=\\frac{1}{3}\operatorname{arctg}\frac{\sqrt{2}}{2}+\frac{\pik}{3};x_{3,4}=\\frac{\pi}{12}+\frac{\pin}{3},k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,957 |
8.060. $\sin 2 x-\sin 3 x+\sin 8 x=\cos \left(7 x+\frac{3 \pi}{2}\right)$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.060. $\sin 2 x-\sin 3 x+\sin 8 x=\cos \left(7 x+\frac{3 \pi}{2}\right)$ | ## Solution.
From the condition
$$
\begin{aligned}
& (\sin 2 x - \sin 3 x) + (\sin 8 x - \sin 7 x) = 0, \Leftrightarrow 2 \sin \frac{2 x - 3 x}{2} \cos \frac{2 x + 3 x}{2} + \\
& + 2 \sin \frac{8 x - 7 x}{2} \cos \frac{8 x + 7 x}{2} = 0, -\sin \frac{x}{2} \cos \frac{5 x}{2} + \sin \frac{x}{2} \cos \frac{15 x}{2} = 0 ... | \frac{\pik}{5},k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,958 |
8.061. $4 \operatorname{tg}^{2} 3 x-\cos ^{-2} 3 x=2$.
Translate to English:
8.061. $4 \tan^{2} 3 x-\cos ^{-2} 3 x=2$. | ## Solution.
Domain of definition: $\cos 3 x \neq 0$.
## We have
$$
\frac{4 \sin ^{2} 3 x}{\cos ^{2} 3 x}-\frac{1}{\cos ^{2} 3 x}-2=0 \Leftrightarrow \frac{4\left(1-\cos ^{2} 3 x\right)}{\cos ^{2} 3 x}-\frac{1}{\cos ^{2} 3 x}-2=0
$$
$$
4\left(1-\cos ^{2} 3 x\right)-1-2 \cos ^{2} 3 x=0, \cos ^{2} 3 x=\frac{1}{2}
$$
... | \\frac{\pi}{12}+\frac{\pik}{3},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,959 |
8.062. $\cos ^{3} x+\cos ^{2} x-4 \cos ^{2} \frac{x}{2}=0$.
8.062. $\cos ^{3} x+\cos ^{2} x-4 \cos ^{2} \frac{x}{2}=0$. | ## Solution.
From the condition
$$
\cos ^{3} x+\cos ^{2} x-4 \cdot \frac{1}{2}(1+\cos x)=0, \cos ^{3} x+\cos ^{2} x-2 \cos x-2=0, \Leftrightarrow
$$
$\Leftrightarrow \cos ^{2} x(\cos x+1)-2(\cos x+1)=0, \Leftrightarrow(\cos x+1)\left(\cos ^{2} x-2\right)=0$.
Then $\cos x+1=0, \cos x=-1, x_{1}=\pi+2 \pi n=\pi(2 n+1)... | \pi(2n+1),\quadn\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,960 |
8.063. $\sin 9 x=2 \sin 3 x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.063. $\sin 9 x=2 \sin 3 x$. | ## Solution.
Rewriting the equation as $\sin 3(3 x)-2 \sin 3 x=0$ and using the formula $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we have
$3 \sin 3 x-4 \sin ^{3} 3 x-2 \sin 3 x=0, 4 \sin ^{3} 3 x-\sin 3 x=0$, $\sin 3 x\left(4 \sin ^{2} 3 x-1\right)=0$.
Then
1) $\sin 3 x=0, 3 x=\pi n, x_{1}=\frac{\pi n}{3}, ... | x_{1}=\frac{\pin}{3};x_{2}=\\frac{\pi}{18}+\frac{\pik}{3},n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,961 |
8.065. $\sin 2z + \cos 2z = \sqrt{2} \sin 3z$. | Solution.
Using the formula $\cos \alpha+\sin \alpha=\sqrt{2} \cos \left(\frac{\pi}{4}-\alpha\right)$, we get
$\sqrt{2} \cos \left(\frac{\pi}{4}-2 z\right)=\sqrt{2} \sin 3 z \Leftrightarrow \cos \left(\frac{\pi}{4}-2 z\right)-\sin 3 z=0 \Leftrightarrow$
$\Leftrightarrow \cos \left(\frac{\pi}{4}-2 z\right)-\cos \left... | z_{1}=\frac{\pi}{20}(8n+3);z_{2}=\frac{\pi}{4}(8k+1),n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,963 |
8.066. $6 \sin ^{2} x+2 \sin ^{2} 2 x=5$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.066. $6 \sin ^{2} x+2 \sin ^{2} 2 x=5$. | Solution.
Rewrite this equation as
$$
\begin{aligned}
& 6 \cdot \frac{1}{2}(1-\cos 2 x)+2\left(1-\cos ^{2} 2 x\right)-5=0 \Leftrightarrow 2 \cos ^{2} 2 x+3 \cos 2 x=0 \\
& \cos 2 x(2 \cos 2 x+3)=0
\end{aligned}
$$
Then
$\cos 2 x=0, 2 x=\frac{\pi}{2}+\pi n, x_{1}=\frac{\pi}{4}+\frac{\pi n}{2}=\frac{\pi}{4}(2 n+1), n... | \frac{\pi}{4}(2n+1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,964 |
8.067. $\sin 3 x+\sin 5 x=2\left(\cos ^{2} 2 x-\sin ^{2} 3 x\right)$. | Solution.
Rewrite the given equation as
$2 \sin \frac{3 x+5 x}{2} \cos \frac{3 x-5 x}{2}=2\left(\frac{1}{2}(1+\cos 4 x)-\frac{1}{2}(1-\cos 6 x)\right) \Leftrightarrow$
$2 \sin 4 x \cos x=\cos 4 x+\cos 6 x \Leftrightarrow$
$\Leftrightarrow 2 \sin 4 x \cos x-2 \cos \frac{4 x+6 x}{2} \cos \frac{4 x-6 x}{2}=0 \Leftrigh... | x_{1}=\frac{\pi}{2}(2n+1),x_{2}=\frac{\pi}{18}(4k+1),n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,965 |
8.068. $\operatorname{tg}\left(\frac{\pi}{2}+x\right)-\operatorname{ctg}^{2} x-\sin ^{-2} x(1+\cos 2 x)=0$.
8.068. $\tan\left(\frac{\pi}{2}+x\right)-\cot^{2} x-\csc ^{2} x(1+\cos 2 x)=0$. | ## Solution.
Domain of definition: $\sin x \neq 0$.
From the condition
$$
\begin{aligned}
& -\operatorname{ctg} x-\operatorname{ctg}^{2} x+\frac{1+2 \cos ^{2} x-1}{\sin ^{2} x}=0,-\operatorname{ctg} x-\operatorname{ctg}^{2} x+2 \operatorname{ctg}^{2} x=0, \\
& \operatorname{ctg}^{2} x-\operatorname{ctg} x=0, \operat... | x_{1}=\frac{\pi}{2}(2k+1);x_{2}=\frac{\pi}{4}(4n+1),\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,966 |
8.069. $2 \sin ^{3} x-\cos 2 x-\sin x=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.069. $2 \sin ^{3} x-\cos 2 x-\sin x=0$. | ## Solution.
We have
$$
\begin{aligned}
& 2 \sin ^{3} x-1+2 \sin ^{2} x-\sin x=0, \quad 2 \sin ^{3} x+2 \sin ^{2} x-\sin x-1=0, \\
& 2 \sin ^{2} x(\sin x+1)-(\sin x+1)=0, \quad(\sin x+1)\left(2 \sin ^{2} x-1\right)=0 .
\end{aligned}
$$
From this:
1) $\sin x+1=0, \quad \sin x=-1, \quad x_{1}=-\frac{\pi}{2}+2 \pi k=\... | x_{1}=\frac{\pi}{2}(4k-1),x_{2}=\frac{\pi}{4}(2n+1)\quadk,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,967 |
8.070. $3 \sin 5 z-2 \cos 5 z=3$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.070. $3 \sin 5 z-2 \cos 5 z=3$. | Solution.
From the condition
$$
3 \sin 2\left(\frac{5}{2} z\right)-2 \cos 2\left(\frac{5}{2} z\right)-3\left(\cos ^{2} \frac{5}{2} z+\sin ^{2} \frac{5}{2} z\right)=0 \Leftrightarrow
$$
$\Leftrightarrow 6 \sin \frac{5}{2} z \cos \frac{5}{2} z-2\left(\cos ^{2} \frac{5}{2} z-\sin ^{2} \frac{5}{2} z\right)-3\left(\cos ^... | z_{1}=\frac{\pi}{10}+\frac{2}{5}\pik;z_{2}=\frac{2}{5}\operatorname{arctg}5+\frac{2}{5}\pin,\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,968 |
8.071. $4 \sin 3 z+\frac{1}{3} \cos 3 z=3$. | ## Solution.
## Rewrite the equation as
$$
\begin{aligned}
& 24 \sin \frac{3 z}{2} \cos \frac{3 z}{2}+\cos ^{2} \frac{3 z}{2}-\sin ^{2} \frac{3 z}{2}-9 \cos ^{2} \frac{3 z}{2}-9 \sin ^{2} \frac{3 z}{2}=0 \Leftrightarrow \\
& \Leftrightarrow 10 \sin ^{2} \frac{3 z}{2}-24 \sin \frac{3 z}{2} \cos \frac{3 z}{2}+8 \cos ^{... | z_{1}=\frac{2}{3}\operatorname{arctg}\frac{2}{5}+\frac{2}{3}\pin;z_{2}=\frac{2}{3}\operatorname{arctg}2+\frac{2}{3}\pik,\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,969 |
8.072. $(\cos 6 x-1) \operatorname{ctg} 3 x=\sin 3 x$.
8.072. $(\cos 6 x-1) \cot 3 x=\sin 3 x$. | ## Solution.
Domain of definition: $\sin 3 x \neq 0$.
We have
$\frac{(\cos 2(3 x)-1) \cos 3 x}{\sin 3 x}-\sin 3 x=0, \quad\left(2 \cos ^{2} 3 x-2\right) \cos 3 x-\sin ^{2} 3 x=0$,
$-2\left(1-\cos ^{2} 3 x\right) \cos 3 x-\sin ^{2} 3 x=0,2 \sin ^{2} 3 x \cos 3 x+\sin ^{2} 3 x=0$,
$\sin ^{2} 3 x(2 \cos 3 x+1)=0$.
S... | \\frac{2}{9}\pi+\frac{2}{3}\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,970 |
8.073. $\frac{\sin \left(\frac{\pi}{2}+x\right) \sin \left(\frac{\pi}{2}-x\right)}{\cos \left(\frac{\pi}{4}+x\right) \cos \left(\frac{\pi}{4}-x\right)}=3$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos \left(\frac{\pi}{4}+x\right) \neq 0, \\ \cos \left(\frac{\pi}{4}-x\right) \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& \cos x \cos x=3 \cos \left(\frac{\pi}{4}+x\right) \cos \left(\frac{\pi}{4}-x\right) \\
& \cos ^{2} x-3 \cos \... | \\frac{\pi}{6}+\pin,\quadn\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,971 |
8.074. $1-\cos (\pi+x)-\sin \frac{3 \pi+x}{2}=0$. | Solution.
We have
$1+\cos x+\cos \frac{x}{2}=0 \Leftrightarrow 1+2 \cos ^{2} \frac{x}{2}-1+\cos \frac{x}{2}=0$,
$2 \cos ^{2} \frac{x}{2}+\cos \frac{x}{2}=0, \quad \cos \frac{x}{2}\left(2 \cos \frac{x}{2}+1\right)=0$.
From this:
1) $\cos \frac{x}{2}=0, \quad \frac{x}{2}=\frac{\pi}{2}+\pi k, \quad x_{1}=\pi+2 \pi k=... | x_1=\pi(2k+ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,972 |
8.075. $9^{\cos x}=9^{\sin x} \cdot 3^{\frac{2}{\cos x}}$.
Domain of definition: $\cos x \neq 0$. | ## Solution.
From the condition
$$
3^{2 \cos x}=3^{2 \sin x} \cdot 3^{\frac{2}{\cos x}} \Leftrightarrow 3^{2 \cos x}=3^{2 \sin x+\frac{2}{\cos x}} \Leftrightarrow 2 \cos x=2 \sin x+\frac{2}{\cos x}
$$
From this,
$\cos ^{2} x-\sin x \cos x-1=0, \sin x \cos x+1-\cos ^{2} x=0$, $\sin x \cos x+\sin ^{2} x=0, \sin x(\co... | x_{1}=\pin;x_{2}=-\frac{\pi}{4}+\pik,n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,973 |
8.076. $\sin x - \sin 2x + \sin 5x + \sin 8x = 0$. | Solution.
We have
$$
\begin{aligned}
& (\sin x - \sin 2x) + (\sin x + \sin 8x) = 0 \Leftrightarrow 2 \sin \frac{x - 2x}{2} \cos \frac{x + 2x}{2} + \\
& + 2 \sin \frac{5x + 8x}{2} \cos \frac{5x - 8x}{2} = 0 \Leftrightarrow -2 \sin \frac{x}{2} \cos \frac{3x}{2} + 2 \sin \frac{13x}{2} \times \\
& \times \cos \frac{3x}{2... | x_1=\frac{\pin}{3};x_2=\frac{\pi}{7}(2l+1)\quadn,\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,974 |
8.077. $2 \sin z - \cos z = \frac{2}{5}$. | Solution.
By transitioning to the half-argument, we find
$$
\begin{aligned}
& 20 \sin \frac{z}{2} \cos \frac{z}{2} - 5 \left( \cos^2 \frac{z}{2} - \sin^2 \frac{z}{2} \right) - 2 \left( \cos^2 \frac{z}{2} + \sin^2 \frac{z}{2} \right) = 0 \\
& 3 \sin^2 \frac{z}{2} + 20 \sin \frac{z}{2} \cos \frac{z}{2} - 7 \cos^2 \frac... | x_1=\frac{\pi}{6}(2k+1);x_2=\frac{\pi}{4}(4n-1),\quadk | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,975 |
8.080. $\cos x - \sqrt{3} \sin x = \cos 3x$. | ## Solution.
From the condition
$$
(\cos x-\cos 3 x)-\sqrt{3} \sin x=0 \Leftrightarrow-2 \sin \frac{x+3 x}{2} \sin \frac{x-3 x}{2}-\sqrt{3} \sin x=0
$$
$2 \sin 2 x \sin x-\sqrt{3} \sin x=0, \sin x(2 \sin 2 x-\sqrt{3})=0$.
From this:
1) $\sin x=0, x_{1}=\pi n, \quad n \in Z$
2) $2 \sin 2 x-\sqrt{3}=0,2 \sin 2 x=\fr... | x_{1}=\pin;x_{2}=(-1)^{k}\frac{\pi}{6}+\frac{\pik}{2},n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,977 |
8.081. $6 \sin ^{2} x+\sin x \cos x-\cos ^{2} x=2$. | Solution.
## Given
$$
\begin{aligned}
& 6 \sin ^{2} x+\sin x \cos x-\cos ^{2} x-2\left(\sin ^{2} x+\cos ^{2} x\right)=0 \\
& 4 \sin ^{2} x+\sin x \cos x-3 \cos ^{2} x=0 \Leftrightarrow 4 \operatorname{tg}^{2} x+\operatorname{tg} x-3=0
\end{aligned}
$$
Solving this equation as a quadratic in $\operatorname{tg} x$, we... | x_{1}=-\frac{\pi}{4}+\pik;x_{2}=\operatorname{arctg}\frac{3}{4}+\pin,k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,978 |
8.082. $\cos 7 x+\sin 8 x=\cos 3 x-\sin 2 x$.
8.082. $\cos 7 x+\sin 8 x=\cos 3 x-\sin 2 x$. | Solution.
We have
$(\cos 7 x-\cos 3 x)+(\sin 8 x+\sin 2 x)=0 \Leftrightarrow-2 \sin \frac{7 x+3 x}{2} \sin \frac{7 x-3 x}{2}+$
$+2 \sin \frac{8 x+2 x}{2} \cos \frac{8 x-2 x}{2}=0,-2 \sin 5 x \sin 2 x+2 \sin 5 x \cos 3 x=0$,
$-2 \sin 5 x(\sin 2 x-\cos 3 x)=0$.
From this:
1) $\sin 5 x=0,5 x=\pi n, x_{1}=\frac{\pi n... | x_{1}=\frac{\pin}{5};x_{2}=\frac{\pi}{2}(4k-1);x_{3}=\frac{\pi}{10}(4+1),n,k,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,979 |
8.083. $\sin ^{2} x-2 \sin x \cos x=3 \cos ^{2} x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.083. $\sin ^{2} x-2 \sin x \cos x=3 \cos ^{2} x$. | Solution.
Dividing both sides of the equation by $\cos ^{2} x \neq 0$, we have $\operatorname{tg}^{2} x-2 \operatorname{tg} x-3=0$.
Solving this equation as a quadratic equation in terms of $\operatorname{tg} x$, we find
$(\operatorname{tg} x)_{1}=-1, x_{1}=-\frac{\pi}{4}+\pi k, k \in Z ;(\operatorname{tg} x)_{2}=3 ... | x_{1}=-\frac{\pi}{4}+\pik,x_{2}=\operatorname{arctg}3+\pin,k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,980 |
8.084. $\cos 5 x+\cos 7 x=\cos (\pi+6 x)$.
8.084. $\cos 5 x+\cos 7 x=\cos (\pi+6 x)$. | Solution.
From the condition
$2 \cos \frac{5 x+7 x}{2} \cos \frac{5 x-7 x}{2}+\cos 6 x=0, 2 \cos 6 x \cos x+\cos 6 x=0$, $\cos 6 x(2 \cos x+1)=0$.
From this:
1) $\cos 6 x=0, 6 x=\frac{\pi}{2}+\pi n, x_{1}=\frac{\pi}{12}+\frac{\pi n}{6}=\frac{\pi}{12}(2 n+1), n \in Z$;
2) $2 \cos x+1=0, \cos x=-\frac{1}{2}, \quad x_... | x_{1}=\frac{\pi}{12}(2n+1),\quadx_{2}=\\frac{2}{3}\pi+2\pik,\quadn,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,981 |
8.085. $4 \sin x \cos \left(\frac{\pi}{2}-x\right)+4 \sin (\pi+x) \cos x+2 \sin \left(\frac{3}{2} \pi-x\right) \cos (\pi+x)=1$. | ## Solution.
By reduction formulas we have
$4 \sin x \sin x-4 \sin x \cos x+2 \cos x \cos x-1=0$
$4 \sin ^{2} x-4 \sin x \cos x+2 \cos ^{2} x-\left(\cos ^{2} x+\sin ^{2} x\right)=0$,
$3 \sin ^{2} x-4 \sin x \cos x+\cos ^{2} x=0, \Leftrightarrow 3 \operatorname{tg}^{2} x-4 \operatorname{tg} x+1=0 ;$
solving the equ... | x_{1}=\operatorname{arctg}\frac{1}{3}+\pik;x_{2}=\frac{\pi}{4}+\pin,k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,982 |
8.086. $\cos 6 x=2 \sin \left(\frac{3 \pi}{2}+2 x\right)$
Translate the above text into English, preserving the original text's line breaks and format, and output the translation result directly.
8.086. $\cos 6 x=2 \sin \left(\frac{3 \pi}{2}+2 x\right)$ | ## Solution.
Representing the equation as $\cos 3(2 x)-2 \sin \left(\frac{3 \pi}{2}+2 x\right)=0$ and applying the formula $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$, we have
$$
\begin{aligned}
& 4 \cos ^{3} 2 x-3 \cos 2 x+2 \cos 2 x=0,4 \cos ^{3} 2 x-\cos 2 x=0, \\
& \cos 2 x\left(4 \cos ^{2} 2 x-1\right)=0 .
... | x_{1}=\frac{\pi}{4}(2k+1),\quadx_{2}=\\frac{\pi}{6}+\frac{\pin}{2},\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,983 |
8.087. $2 \sin x \cos \left(\frac{3 \pi}{2}+x\right)-3 \sin (\pi-x) \cos x+\sin \left(\frac{\pi}{2}+x\right) \cos x=0$. | ## Solution.
By reduction formulas
$$
\begin{aligned}
& 2 \sin x \sin x-3 \sin x \cos x+\cos x \cos x=0 \\
& 2 \sin ^{2} x-3 \sin x \cos x+\cos ^{2} x=0 \Leftrightarrow 2 \operatorname{tg}^{2} x-3 \operatorname{tg} x+1=0
\end{aligned}
$$
solving the equation as a quadratic in $\operatorname{tg} x$, we find $(\operat... | x_{1}=\operatorname{arctg}\frac{1}{2}+\pik;x_{2}=\frac{\pi}{4}(4n+1)\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,984 |
8.089. $\cos \left(2 t-18^{\circ}\right) \operatorname{tg} 50^{\circ}+\sin \left(2 t-18^{\circ}\right)=\frac{1}{2 \cos 130^{\circ}}$. | ## Solution.
We have
$$
\begin{aligned}
& \frac{\cos \left(2 t-18^{\circ}\right) \sin 50^{\circ}}{\cos 50^{\circ}}+\sin \left(2 t-18^{\circ}\right)=\frac{1}{2 \cos \left(180^{\circ}-50^{\circ}\right)} \Leftrightarrow \\
& \Leftrightarrow \frac{\cos \left(2 t-18^{\circ}\right) \sin 50^{\circ}+\sin \left(2 t-18^{\circ}... | t_{1}=-31+180k;t_{2}=89+180k,k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,986 |
8.090. $\operatorname{tg} \frac{t}{2} \operatorname{ctg} \frac{3 t}{2}+\cos ^{-1} \frac{t}{2} \sin ^{-1} \frac{3 t}{2}=1$.
8.090. $\tan \frac{t}{2} \cot \frac{3 t}{2}+\cos ^{-1} \frac{t}{2} \sin ^{-1} \frac{3 t}{2}=1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos \frac{t}{2} \neq 0, \\ \sin \frac{3 t}{2} \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& \frac{\sin \frac{t}{2} \cos \frac{3 t}{2}}{\cos \frac{t}{2} \sin \frac{3 t}{2}}+\frac{1}{\cos \frac{t}{2} \sin \frac{3 t}{2}}-1=0 \Rightar... | \frac{\pi}{2}(4k+1),\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,987 |
8.094. $\frac{1-\cos x}{\sin \frac{x}{2}}=2$. | ## Solution.
Domain of definition: $\sin \frac{x}{2} \neq 0$.
From the condition
$$
\begin{aligned}
& \frac{1-\cos 2\left(\frac{x}{2}\right)}{\sin \frac{x}{2}}=2 \Leftrightarrow \frac{1-1+2 \sin ^{2} \frac{x}{2}}{\sin \frac{x}{2}}=2 \Rightarrow \sin \frac{x}{2}=1, \frac{x}{2}=\frac{\pi}{2}+2 \pi k \\
& x=\pi+4 \pi k... | \pi(4k+1),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,988 |
8.095. $\sin \frac{7 x}{2} \cos \frac{3 x}{2}+\sin \frac{x}{2} \cos \frac{5 x}{2}+\sin 2 x \cos 7 x=0$. | ## Solution.
Applying the formula $\sin \alpha \cos \beta=\frac{1}{2}(\sin (\alpha-\beta)+\sin (\alpha+\beta))$, we write the equation as
$$
\begin{aligned}
& \frac{1}{2}\left(\sin \left(\frac{7 x}{2}-\frac{3 x}{2}\right)+\sin \left(\frac{7 x}{2}+\frac{3 x}{2}\right)\right)+\frac{1}{2}\left(\sin \left(\frac{x}{2}-\fr... | \frac{\pin}{6},n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,989 |
8.096. $\sin 3 x+\sin 5 x=\sin 4 x$.
8.096. $\sin 3 x+\sin 5 x=\sin 4 x$. | Solution.
Rewrite the given equation as
$$
\begin{aligned}
& 2 \sin \frac{3 x+5 x}{2} \cos \frac{3 x-5 x}{2}-\sin 4 x=0,2 \sin 4 x \cos x-\sin 4 x=0 \\
& \sin 4 x(2 \cos x-1)=0
\end{aligned}
$$
From this:
1) $\sin 4 x=0,4 x=\pi n, x_{1}=\frac{\pi n}{4}, n \in Z$;
2) $2 \cos x-1=0, \quad \cos x=\frac{1}{2}, \quad x_... | x_{1}=\frac{\pin}{4};x_{2}=\frac{\pi}{3}(6k\1),n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,990 |
8.097. $\sin z-\sin ^{2} z=\cos ^{2} z-\cos z$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.097. $\sin z-\sin ^{2} z=\cos ^{2} z-\cos z$. | Solution.
$\sin z+\cos z-\left(\sin ^{2} z+\cos ^{2} z\right)=0, \quad \sin z+\cos z-1=0$, $\sin 2\left(\frac{z}{2}\right)+\cos 2\left(\frac{z}{2}\right)-\left(\sin ^{2} \frac{z}{2}+\cos ^{2} \frac{z}{2}\right)=0, \Leftrightarrow 2 \sin \frac{z}{2} \cos \frac{z}{2}+\cos ^{2} \frac{z}{2}-$ $-\sin ^{2} \frac{z}{2}-\sin ... | z_{1}=2\pin;z_{2}=\frac{\pi}{2}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,991 |
8.098. $\sin z+\sin 2 z+\sin 3 z=\cos z+\cos 2 z+\cos 3 z$.
8.098. $\sin z + \sin 2z + \sin 3z = \cos z + \cos 2z + \cos 3z$. | Solution.
Rewrite the given equation as
$$
\begin{aligned}
& 2 \sin \frac{z+3 z}{2} \cos \frac{z-3 z}{2}+\sin 2 z=2 \cos \frac{z+3 z}{2} \cos \frac{z-3 z}{2}+\cos 2 z \Leftrightarrow \\
& \Leftrightarrow 2 \sin 2 z \cos z+\sin 2 z=2 \cos 2 z \cos z+\cos 2 z, \sin 2 z(2 \cos z+1)- \\
& -\cos 2 z(2 \cos z+1)=0,(2 \cos ... | z_{1}=\frac{2}{3}\pi(3k\1)\quadz_{2}=\frac{\pi}{8}(4n+1),\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,992 |
8.099. $\operatorname{ctg} x-\operatorname{tg} x+2 \cdot\left(\frac{1}{\operatorname{tg} x+1}+\frac{1}{\operatorname{tg} x-1}\right)=4$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\operatorname{tg} x \neq \pm 1, \\ \cos x \neq 0, \\ \sin x \neq 0\end{array}\right.$
From the condition
$\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}+\frac{\frac{4 \sin x}{\cos x}}{\frac{\sin ^{2} x}{\cos ^{2} x}-1}=4 \Leftrightarrow \frac{\cos ^{2} x-\sin ^{2}... | \frac{\pi}{16}(4n+1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,993 |
8.100. $1-\cos 6 x=\tan 3 x$. | ## Solution.
Domain of definition: $\cos 3 x \neq 0, 3 x \neq \frac{\pi}{2}+\pi n, x \neq \frac{\pi}{6}(2 n+1), n \in Z$.
## We have
$1-\cos 6 x-\frac{\sin 6 x}{1+\cos 6 x}=0, \quad(1-\cos 6 x)(1+\cos 6 x)-\sin 6 x=0$,
$1-\cos ^{2} 6 x-\sin 6 x=0, \sin ^{2} 6 x-\sin 6 x=0, \sin 6 x(\sin 6 x-1)=0$.
Then:
1) $\sin ... | x_{1}=\frac{\pi}{3};x_{2}=\frac{\pi}{12}(4k+1),,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,994 |
8.101. $\sqrt{2} \cos x+\cos 2 x+\cos 4 x=0$. | Solution.
Rewrite the equation as
$$
\begin{aligned}
& \sqrt{2} \cos x+2 \cos \frac{2 x+4 x}{2} \cos \frac{2 x-4 x}{2}=0 \Leftrightarrow \sqrt{2} \cos x+ \\
& +2 \cos 3 x \cos x=0 \Leftrightarrow \cos x(\sqrt{2}+2 \cos 3 x)=0
\end{aligned}
$$
From this,
1) $\cos x=0, x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), ... | x_{1}=\frac{\pi}{2}(2k+1);x_{2}=\frac{\pi}{12}(8n\3),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,995 |
8.102. $\sin ^{4} x+\cos ^{4} x=\sin 2 x-0.5$. | Solution.
We have
$$
\begin{aligned}
& \left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x-\sin 2 x+0.5=0 \Leftrightarrow 1-\frac{1}{2} \sin ^{2} 2 x- \\
& -\sin 2 x+\frac{1}{2}=0, \sin ^{2} 2 x+2 \sin 2 x-3=0
\end{aligned}
$$
Solving this equation as a quadratic equation in terms of $\sin 2 x$, we ge... | \frac{\pi}{4}(4n+1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,996 |
8.103. $2 \cos 2 x+2 \tan^{2} x=5$. | ## Solution.
Domain of definition: $\cos x \neq 0$.
Rewrite the equation as
$$
\begin{aligned}
& 2 \cos 2 x+\frac{2(1-\cos 2 x)}{1+\cos 2 x}-5=0 \Rightarrow 2 \cos 2 x(1+\cos 2 x)+2(1-\cos 2 x)- \\
& -5(1+\cos 2 x)=0,2 \cos ^{2} 2 x-5 \cos 2 x-3=0 .
\end{aligned}
$$
Solving this equation as a quadratic in $\cos 2 x... | \frac{\pi}{3}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,997 |
8.105. $\sin ^{4} 2 x+\cos ^{4} 2 x=\sin 2 x \cos 2 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.105. $\sin ^{4} 2 x+\cos ^{4} 2 x=\sin 2 x \cos 2 x$. | Solution.
We have
$$
\begin{aligned}
& \left(\sin ^{2} 2 x+\cos ^{2} 2 x\right)^{2}-2 \sin ^{2} 2 x \cos ^{2} 2 x-\sin 2 x \cos 2 x=0 \\
& 1-2 \sin ^{2} 2 x \cos ^{2} 2 x-\sin 2 x \cos 2 x=0 \Leftrightarrow \sin ^{2} 4 x+\sin 4 x-2=0
\end{aligned}
$$
Solving this equation as a quadratic in $\sin 4 x$, we find $\sin ... | \frac{\pi}{8}(4n+1),\quadn\inZ | Algebra | proof | Yes | Yes | olympiads | false | 50,998 |
8.106. $\cos \left(3 x-30^{\circ}\right)-\sin \left(3 x-30^{\circ}\right) \tan 30^{\circ}=\frac{1}{2 \cos 210^{\circ}}$. | ## Solution.
From the condition
$$
\begin{aligned}
& \cos \left(3 x-30^{\circ}\right)-\frac{\sin \left(3 x-30^{\circ}\right) \sin 30^{\circ}}{\cos 30^{\circ}}=\frac{1}{2 \cos \left(180^{\circ}+30^{\circ}\right)} \Leftrightarrow \\
& \frac{\cos \left(3 x-30^{\circ}\right) \cos 30^{\circ}-\sin \left(3 x-30^{\circ}\righ... | \\frac{2\pi}{9}+\frac{2}{3}\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,999 |
8.107. $4 \sin x + \cos x = 4$. | Solution.
Rewrite the equation as
$$
\begin{aligned}
& 4 \sin 2\left(\frac{x}{2}\right)+\cos 2\left(\frac{x}{2}\right)-4\left(\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}\right)=0 \Leftrightarrow 8 \sin \frac{x}{2} \cos \frac{x}{2}+ \\
& +\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}-4 \cos ^{2} \frac{x}{2}-4 \sin ^{2} ... | x_{1}=2\operatorname{arctg}\frac{3}{5}+2\pin;x_{2}=\frac{\pi}{2}(4k+1),\quadn,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,000 |
8.108. $2 \sin ^{2} z+\tan ^{2} z=2$. | ## Solution.
Domain of definition: $\cos z \neq 0$.
From the condition
$$
\begin{aligned}
& 2 \sin ^{2} z+\frac{\sin ^{2} z}{\cos ^{2} z}-2=0 \Leftrightarrow 2 \sin ^{2} z+\frac{\sin ^{2} z}{1-\sin ^{2} z}-2=0 \Rightarrow \\
& \Rightarrow 2 \sin ^{2} z\left(1-\sin ^{2} z\right)+\sin ^{2} z-2\left(1-\sin ^{2} z\right... | \frac{\pi}{4}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,001 |
8.109. $\cos 2 x+\cos 6 x+2 \sin ^{2} x=1$. | ## Solution.
From the condition
$\cos 2 x+\cos 6 x-\left(1-2 \sin ^{2} x\right)=0 \Leftrightarrow \cos 2 x+\cos 6 x-\cos 2 x=0, \cos 6 x=0$.
Then $6 x=\frac{\pi}{2}+\pi n, x=\frac{\pi}{12}+\frac{\pi n}{6}=\frac{\pi}{12}(2 n+1), n \in Z$.
Answer: $\quad x=\frac{\pi}{12}(2 n+1), \quad n \in Z$. | \frac{\pi}{12}(2n+1),\quadn\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,002 |
8.110. $\cos 3 x \cos 6 x=\cos 4 x \cos 7 x$.
8.110. $\cos 3 x \cos 6 x=\cos 4 x \cos 7 x$. | ## Solution.
We have
$$
\begin{aligned}
& \frac{1}{2}(\cos (3 x-6 x)+\cos (3 x+6 x))=\frac{1}{2}(\cos (4 x-7 x)+\cos (4 x+7 x)) \\
& \cos 3 x+\cos 9 x-\cos 3 x-\cos 11 x=0, \cos 9 x-\cos 11 x=0 \Leftrightarrow \\
& \Leftrightarrow-2 \sin \frac{9 x+11 x}{2} \sin \frac{9 x-11 x}{2}=0, \quad \sin 10 x \sin x=0
\end{alig... | \frac{\pin}{10},\quadn\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,003 |
8.112. $\operatorname{ctg}^{3} x+\sin ^{-2} x-3 \operatorname{ctg} x-4=0$.
8.112. $\cot^{3} x+\csc ^{2} x-3 \cot x-4=0$. | ## Solution.
Domain of definition: $\quad \sin x \neq 0$.
## From the condition
$$
\frac{\cos ^{3} x}{\sin ^{3} x}+\frac{1}{\sin ^{2} x}-\frac{3 \cos x}{\sin x}-4=0 \Rightarrow \cos ^{3} x+\sin x-3 \cos x \sin ^{2} x-4 \sin ^{3} x=0
$$
$\cos ^{3} x+\sin x\left(\sin ^{2} x+\cos ^{2} x\right)-3 \cos x \sin ^{2} x-4 \... | x_{1}=\frac{3\pi}{4}+\pin;x_{2}=\\frac{\pi}{6}+\pik,n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,004 |
8.114. $1+\sin x-\cos 5 x-\sin 7 x=2 \cos ^{2} \frac{3}{2} x$.
8.114. $1+\sin x-\cos 5x-\sin 7x=2 \cos^2 \frac{3}{2}x$. | ## Solution.
Let's rewrite the given equation as
$1+\sin x-\cos 5 x-\sin 7 x=1+\cos 3 x \Leftrightarrow (\sin x-\sin 7 x)-(\cos 5 x+\cos 3 x)=0 \Leftrightarrow$ $\Leftrightarrow -2 \sin 3 x \cos 4 x - 2 \cos 4 x \cos x = 0, \quad -2 \cos 4 x (\sin 3 x + \cos x) = 0$.
From this, we have:
1) $\cos 4 x = 0, 4 x = \fra... | x_{1}=\frac{\pi}{8}(2k+1),x_{2}=\frac{\pi}{4}(4n-1)\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,005 |
8.115. $\frac{\sin z}{1+\cos z}=2-\operatorname{ctg} z$.
8.115. $\frac{\sin z}{1+\cos z}=2-\cot z$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\sin z \neq 0, \\ \cos z \neq-1 .\end{array}\right.$
From the condition
$\frac{\sin z}{1+\cos z}-2+\frac{\cos z}{\sin z}=0 \Rightarrow \sin ^{2} z-2 \sin z(1+\cos z)+\cos z(1+\cos z)=0$, $\sin ^{2} z-2 \sin z-2 \sin z \cos z+\cos z+\cos ^{2} z=0,(1-2 \sin z)+$ ... | (-1)^{k}\frac{\pi}{6}+\pik,\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,006 |
8.116. $\sin \left(15^{\circ}+x\right)+\cos \left(45^{\circ}+x\right)+\frac{1}{2}=0$. | ## Solution.
We have
$$
\begin{aligned}
& \sin \left(15^{\circ}+x\right)+\sin \left(90^{\circ}-45^{\circ}-x\right)+\frac{1}{2}=0, \sin \left(15^{\circ}+x\right)+\sin \left(45^{\circ}-x\right)+ \\
& +\frac{1}{2}=0 \Leftrightarrow 2 \sin \frac{15^{\circ}+x+45^{\circ}-x}{2} \cos \frac{15^{\circ}+x-45^{\circ}+x}{2}+\frac... | x_{1}=-105+360k;x_{2}=135+360k,\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,007 |
8.117. $1+\sin 2x=\sin x+\cos x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
8.117. $1+\sin 2x=\sin x+\cos x$. | ## Solution.
Let's rewrite the equation as
$\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x-(\sin x+\cos x)=0, \quad(\sin x+\cos x)^{2}-$ $-(\sin x+\cos x)=0, \quad(\sin x+\cos x)(\sin x+\cos x-1)=0$.
From this, either $\sin x+\cos x=0$, or $\sin x+\cos x-1=0$. From the first equation, $\operatorname{tg} x=-1, x_{1}=-\frac{... | x_{1}=\frac{\pi}{4}(4k-1),x_{2}=2\pin,x_{3}=\frac{\pi}{2}(4+1),k,n,\inZ | Other | math-word-problem | Yes | Yes | olympiads | false | 51,008 |
8.118. $3(1-\sin t)+\sin ^{4} t=1+\cos ^{4} t$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
8.118. $3(1-\sin t)+\sin ^{4} t=1+\cos ^{4} t$. | ## Solution.
From the condition
$$
\begin{aligned}
& 3-3 \sin t+\sin ^{4} t-1-\left(\cos ^{2} t\right)^{2}=0, \sin ^{4} t-3 \sin t+2-\left(1-\sin ^{2} t\right)^{2}=0 \Leftrightarrow \\
& \Leftrightarrow 2 \sin ^{2} t-3 \sin t+1=0
\end{aligned}
$$
Solving the last equation as a quadratic equation in terms of $\sin t$... | t_{1}=(-1)^{k}\frac{\pi}{6}+\pik;t_{2}=\frac{\pi}{2}(4n+1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,009 |
8.119. $\operatorname{tg}\left(\frac{5 \pi}{2}+x\right)-3 \operatorname{tg}^{2} x=(\cos 2 x-1) \cos ^{-2} x$.
8.119. $\tan\left(\frac{5 \pi}{2}+x\right)-3 \tan^{2} x=(\cos 2 x-1) \cos^{-2} x$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0 \\ \sin x \neq 0\end{array}\right.$
## We have
$$
\begin{aligned}
& -\operatorname{ctg} x-3 \operatorname{tg}^{2} x=\frac{\cos 2 x-1}{\cos ^{2} x} \Leftrightarrow \frac{1}{\operatorname{tg} x}+3 \operatorname{tg}^{2} x=\frac{1-\cos 2 x}{\frac{1... | \frac{\pi}{4}(4k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,010 |
8.120. $\cos ^{2} \frac{x}{2}+\cos ^{2} \frac{3 x}{2}-\sin ^{2} 2 x-\sin ^{2} 4 x=0$. | ## Solution.
By the formulas for reducing the degree
$\frac{1}{2}(1+\cos x)+\frac{1}{2}(1+\cos 3 x)-\frac{1}{2}(1-\cos 4 x)-\frac{1}{2}(1-\cos 8 x)=0$,
$(\cos x+\cos 3 x)+(\cos 4 x+\cos 8 x)=0 \Leftrightarrow 2 \cos 2 x \cos x+2 \cos 6 x \cos 2 x=0$, $2 \cos 2 x(\cos x+\cos 6 x)=0$.
From this:
1) $\cos 2 x=0, 2 x=... | x_{1}=\frac{\pi}{4}(2k+1),x_{2}=\frac{\pi}{7}(2n+1),x_{3}=\frac{\pi}{5}(2+1)\quadk,n,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,011 |
8.121. $\frac{\sin ^{2} x-2}{\sin ^{2} x-4 \cos ^{2} \frac{x}{2}}=\tan ^{2} \frac{x}{2}$. | ## Solution.
DZ: $\left\{\begin{array}{l}\cos ^{2} \frac{x}{2} \neq 0, \\ \sin ^{2} \frac{x}{2}-4 \cos ^{2} \frac{x}{2} \neq 0 .\end{array}\right.$
Rewrite the given equation as
$$
\begin{aligned}
& \frac{\sin ^{2} x-2}{\sin ^{2} x-2(1+\cos x)}=\frac{1-\cos x}{1+\cos x}, \frac{1-\cos ^{2} x-2}{1-\cos ^{2} x-2-2 \cos... | \frac{\pi}{2}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,012 |
8.122. $\cos ^{2} x+\cos ^{2} 2 x-\cos ^{2} 3 x-\cos ^{2} 4 x=0$. | ## Solution.
By the formulas for reducing the power
$$
\begin{aligned}
& \frac{1}{2}(1+\cos 2 x)+\frac{1}{2}(1+\cos 4 x)-\frac{1}{2}(1+\cos 6 x)-\frac{1}{2}(1+\cos 8 x)=0 \\
& (\cos 2 x+\cos 4 x)-(\cos 6 x+\cos 8 x)=0 \Leftrightarrow 2 \cos 3 x \cos x-2 \cos 7 x \cos x=0 \\
& 2 \cos x(\cos 3 x-\cos 7 x)=0
\end{aligne... | x_{1}=\frac{\pin}{5};x_{2}=\frac{\pi}{2},n,\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,013 |
8.123. $\sin 3 x-4 \sin x \cos 2 x=0$. | ## Solution.
Let's rewrite the equation as
$\sin 3 x-2(\sin (x-2 x)+\sin (x+2 x))=0, \quad \sin 3 x+2 \sin x-2 \sin 3 x=0$, $\sin 3 x-2 \sin x=0$.
Since $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we have $3 \sin x-4 \sin ^{3} x-2 \sin x=0$, $4 \sin ^{3} x-\sin x=0, \sin x\left(4 \sin ^{2} x-1\right)=0$.
From... | x_{1}=\pin;x_{2}=\frac{\pi}{6}(6k\1)\quadn,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,014 |
8.124. $\operatorname{tg} x+\operatorname{ctg} x=2 \cos^{-1} 4 x$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0, \\ \cos 4 x \neq 0 .\end{array}\right.$
From the condition
$$
\begin{aligned}
& \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}-\frac{2}{\cos 4 x}=0, \frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}-\frac{2}{\cos 4 x}=0 \Leftrightar... | \frac{\pi}{12}(4k+1),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,015 |
8.126. $\frac{1}{1+\cos ^{2} z}+\frac{1}{1+\sin ^{2} z}=\frac{16}{11}$. | Solution.
Rewrite the given equation as
$$
\begin{aligned}
& \frac{1}{1+\frac{1}{2}(1+\cos 2 z)}+\frac{1}{1+\frac{1}{2}(1-\cos 2 z)}-\frac{16}{11}=0, \frac{2}{3+\cos 2 z}+\frac{2}{3-\cos 2 z}-\frac{16}{11}=0 \Rightarrow \\
& \Rightarrow \cos ^{2} 2 z=\frac{3}{4}, \cos 2 z= \pm \frac{\sqrt{3}}{2}, 2 z= \pm \frac{\pi}{... | \frac{\pi}{12}(6k\1),\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,016 |
8.128. $\cos 4 x \cos (\pi+2 x)-\sin 2 x \cos \left(\frac{\pi}{2}-4 x\right)=\frac{\sqrt{2}}{2} \sin 4 x$. | ## Solution.
By reduction formulas
$-\cos 4 x \cos 2 x-\sin 2 x \sin 4 x-\frac{\sqrt{2}}{2} \cdot 2 \sin 2 x \cos 2 x=0, \cos 4 x \cos 2 x+$ $+\sin 4 x \sin 2 x+\sqrt{2} \sin 2 x \cos 2 x=0 \Leftrightarrow \cos 2 x+\sqrt{2} \sin 2 x \cos 2 x=0$, $\cos 2 x(1+\sqrt{2} \sin 2 x)=0$.
From this:
1) $\cos 2 x=0, 2 x=\fra... | x_{1}=\frac{\pi}{4}(2n+1);x_{2}=(-1)^{k+1}\frac{\pi}{8}+\frac{\pik}{2},n,k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 51,017 |
8.129. $\sin x-\sin 3 x-\sin 5 x+\sin 7 x=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.129. $\sin x-\sin 3 x-\sin 5 x+\sin 7 x=0$. | Solution.
We have
$$
\begin{aligned}
& (\sin x+\sin 7 x)-(\sin 3 x+\sin 5 x)=0 \Leftrightarrow 2 \sin \frac{x+7 x}{2} \cos \frac{x-7 x}{2}- \\
& -2 \sin \frac{3 x+5 x}{2} \cos \frac{3 x-5 x}{2}=0,2 \sin 4 x \cos 3 x-2 \sin 4 x \cos x=0 \\
& 2 \sin 4 x(\cos 3 x-\cos x)=0
\end{aligned}
$$
From this:
1) $\sin 4 x=0,4 ... | \frac{\pik}{4},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,018 |
8.130. $\sin 3 x-\sin 7 x=\sqrt{3} \sin 2 x$. | ## Solution.
Let's rewrite the given equation as
$2 \sin \frac{3 x-7 x}{2} \cos \frac{3 x+7 x}{2}-\sqrt{3} \sin 2 x=0, -2 \sin 2 x \cos 5 x-\sqrt{3} \sin 2 x=0$, $-\sin 2 x(2 \cos 5 x+\sqrt{3})=0$
From this:
1) $\sin 2 x=0, 2 x=\pi k, x_{1}=\frac{\pi k}{2}, k \in Z$
2) $2 \cos 5 x+\sqrt{3}=0, \quad \cos 5 x=-\frac{... | x_{1}=\frac{\pik}{2};x_{2}=\\frac{\pi}{6}+\frac{2}{5}\pin,\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,019 |
8.131. $\sqrt{3}-\tan x=\tan\left(\frac{\pi}{3}-x\right)$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos \left(\frac{\pi}{3}-x\right) \neq 0 .\end{array}\right.$
From the condition
$$
\sqrt{3}=\operatorname{tg} x+\operatorname{tg}\left(\frac{\pi}{3}-x\right) \Leftrightarrow \sqrt{3}=\frac{\sin \left(x+\frac{\pi}{3}-x\right)}{\cos x \cos \... | x_{1}=\pik;x_{2}=\frac{\pi}{3}(3+1)\quadk,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,020 |
8.133. $\sin ^{2} 3 x+\sin ^{2} 4 x=\sin ^{2} 5 x+\sin ^{2} 6 x$.
8.133. $\sin ^{2} 3 x+\sin ^{2} 4 x=\sin ^{2} 5 x+\sin ^{2} 6 x$.
The equation remains the same in English as it is a mathematical expression. | ## Solution.
Using the formulas for reducing the power
$$
\begin{aligned}
& \frac{1}{2}(1-\cos 6 x)+\frac{1}{2}(1-\cos 8 x)=\frac{1}{2}(1-\cos 10 x)+\frac{1}{2}(1-\cos 12 x) \\
& (\cos 6 x+\cos 8 x)-(\cos 10 x+\cos 12 x)=0 \Leftrightarrow 2 \cos \frac{6 x+8 x}{2} \cos \frac{6 x-8 x}{2}- \\
& -2 \cos \frac{10 x+12 x}{... | x_{1}=\frac{\pi}{2};x_{2}=\frac{\pin}{9},,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,021 |
8.135. $\sin ^{4} x+\cos ^{4} x=\cos ^{2} 2 x+0.25$. | ## Solution.
From the condition
$$
\begin{aligned}
& \left(\sin ^{2} x\right)^{2}+\left(\cos ^{2} x\right)^{2}=\cos ^{2} 2 x+0.25 \Leftrightarrow\left(\frac{1-\cos 2 x}{2}\right)^{2}+\left(\frac{1+\cos 2 x}{2}\right)^{2}= \\
& =\cos ^{2} 2 x+0.25 \Leftrightarrow 2 \cos ^{2} 2 x-1=0 \Leftrightarrow \cos 4 x=0,4 x=\fra... | \frac{\pi}{8}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,023 |
### 8.136. $\sin 2 z-4 \cos 2 z=4$. | ## Solution.
Let's rewrite the given equation as
$$
2 \sin z \cos z-4\left(\cos ^{2} z-\sin ^{2} z\right)-4\left(\cos ^{2} z+\sin ^{2} z\right)=0
$$
$$
2 \sin z \cos z-8 \cos ^{2} z=0, \quad 2 \cos z(\sin z-4 \cos z)=0
$$
## From this:
1) $\cos z=0, \quad z_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$
2)... | z_{1}=\frac{\pi}{2}(2k+1),z_{2}=\operatorname{arctg}4+\pin,k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,024 |
8.138. $\sin ^{2}\left(\frac{\pi}{8}+t\right)=\sin t+\sin ^{2}\left(\frac{\pi}{8}-t\right)$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.138. $\sin ^{2}\left(\frac{\pi}{8}+t\right)=\sin t+\sin ^{2}\left(\frac{\pi}{8}-t\right)$ | ## Solution.
By the formulas for reducing the power
$$
\begin{aligned}
& \frac{1}{2}\left(1-\cos \left(\frac{\pi}{4}+2 t\right)\right)=\sin t+\frac{1}{2}\left(1-\cos \left(\frac{\pi}{4}-2 t\right)\right) 2 \sin t+\cos \left(\frac{\pi}{4}+2 t\right)- \\
& -\cos \left(\frac{\pi}{4}-2 t\right)=0 \Leftrightarrow 2 \sin t... | t_{1}=\pin;t_{2}=\frac{\pi}{4}(8k\1),n,k\in\mathbb{Z} | Algebra | proof | Yes | Yes | olympiads | false | 51,025 |
8.139. $\sin ^{3} \frac{x}{3}-\sin ^{2} \frac{x}{3} \cos \frac{x}{3}-3 \sin \frac{x}{3} \cos ^{2} \frac{x}{3}+3 \cos ^{3} \frac{x}{3}=0$. | ## Solution.
Let's rewrite the equation as
$$
\begin{aligned}
& \sin ^{2} \frac{x}{3}\left(\sin \frac{x}{3}-\cos \frac{x}{3}\right)-3 \cos ^{2} \frac{x}{3}\left(\sin \frac{x}{3}-\cos \frac{x}{3}\right)=0 \\
& \left(\sin \frac{x}{3}-\cos \frac{x}{3}\right)\left(\sin ^{2} \frac{x}{3}-3 \cos ^{2} \frac{x}{3}\right)=0
\e... | x_{1}=\frac{3\pi}{4}(4k+1);x_{2}=\pi(3n\1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,026 |
8.140. $\operatorname{tg}\left(x-15^{\circ}\right) \operatorname{ctg}\left(x+15^{\circ}\right)=\frac{1}{3}$. | ## Solution.
We have
$$
\frac{\sin \left(x-15^{\circ}\right) \cos \left(x+15^{\circ}\right)}{\cos \left(x-15^{\circ}\right) \sin \left(x+15^{\circ}\right)}-\frac{1}{3}=0 \Leftrightarrow \frac{\sin \left(x-15^{\circ}-x-15^{\circ}\right)+\sin \left(x-15^{\circ}+x+15^{\circ}\right)}{\sin \left(x+15^{\circ}-x+15^{\circ}\... | 45(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,027 |
8.141. $\cos (x+1) \sin 2(x+1)=\cos 3(x+1) \sin 4(x+1)$ | ## Solution.
Rewrite the equation as
$$
\begin{aligned}
& \frac{1}{2}(\sin (2 x+2-x-1)+\sin (2 x+2+x+1))- \\
& -\frac{1}{2}(\sin (4 x+4-3 x-3)+\sin (4 x+4+3 x+3))=0, \quad \sin (x+1)+\sin (x+3)- \\
& -\sin (x+1)-\sin (7 x+7)=0, \sin (3 x+3)-\sin (7 x+7)=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos \frac{3 x+3+7 x+7}... | x_{1}=-1+\frac{\pi}{10}(2k+1),\quadx_{2}=-1+\frac{\pin}{2},\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,028 |
8.142. $\cos (4 x+2)+3 \sin (2 x+1)=2$. | ## Solution.
From the condition
$$
\begin{aligned}
& \cos 2(2 x+1)+3 \sin (2 x+1)-2=0 \Leftrightarrow 1-2 \sin ^{2}(2 x+1)+3 \sin (2 x+1)-2=0, \\
& 2 \sin ^{2}(2 x+1)-3 \sin (2 x+1)+1=0 .
\end{aligned}
$$
Solving this equation as a quadratic in $\sin (2 x+1)$, we find
$$
(\sin (2 x+1))_{1}=\frac{1}{2}, 2 x_{1}+1=(-... | x_{1}=(-1)^{k}\frac{\pi}{12}-\frac{1}{2}+\frac{\pik}{2};x_{2}=\frac{\pi}{4}(4n+1)-\frac{1}{2},k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,029 |
8.143. $\cos 4 x+2 \cos ^{2} x=1$.
8.143. $\cos 4 x+2 \cos ^{2} x=1$. | ## Solution.
We have
$$
\begin{aligned}
& \cos 2(2 x)+2 \cos ^{2} x-1=0, \Leftrightarrow 2 \cos ^{2} 2 x-1+\cos 2 x=0 \\
& 2 \cos ^{2} 2 x+\cos 2 x-1=0
\end{aligned}
$$
Solving this equation as a quadratic in terms of $\cos 2 x$, we find $(\cos 2 x)_{1}=-1, 2 x_{1}=\pi+2 \pi k, x_{1}=\frac{\pi}{2}+\pi k, k \in \math... | \frac{\pi}{6}(2k+1)\quadk\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,030 |
8.145. $\cos x - \cos 2x = \sin 3x$. | Solution.
Rewrite the equation as
$$
\begin{aligned}
& 2 \sin \frac{x+2 x}{2} \sin \frac{2 x-x}{2}-\sin 3 x=0, \quad 2 \sin \frac{3 x}{2} \sin \frac{x}{2}-\sin 2\left(\frac{3 x}{2}\right)=0 \\
& 2 \sin \frac{3 x}{2} \sin \frac{x}{2}-2 \sin \frac{3 x}{2} \cos \frac{3 x}{2}=0, \quad 2 \sin \frac{3 x}{2}\left(\sin \frac... | x_{1}=\frac{2}{3}\pik;x_{2}=\frac{\pi}{4}(4+1);x_{3}=\frac{\pi}{2}(4n-1),\quadk,,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,031 |
8.146. $\operatorname{tg} x+\operatorname{tg} 50^{\circ}+\operatorname{tg} 70^{\circ}=\operatorname{tg} x \operatorname{tg} 50^{\circ} \operatorname{tg} 70^{\circ}$. | ## Solution.
Domain of definition: $\cos x \neq 0$.
From the condition
$\frac{\sin x}{\cos x}+\frac{\sin 50^{\circ}}{\cos 50^{\circ}}+\frac{\sin 70^{\circ}}{\cos 70^{\circ}}-\frac{\sin x \sin 50^{\circ} \sin 70^{\circ}}{\cos x \cos 50^{\circ} \cos 70^{\circ}}=0 \Rightarrow$
$\Rightarrow\left(\sin x \cos 50^{\circ} ... | 60+180n,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,032 |
8.147. $\cos x - \sin x = 4 \cos x \sin^2 x$. | Solution.
By the formulas for reducing the degree
$\cos x-\sin x=2 \cos x(1-\cos 2 x) \Leftrightarrow 2 \cos x \cos 2 x-(\cos x+\sin x)=0$, $2 \cos x(\cos x+\sin x)(\cos x-\sin x)-(\cos x+\sin x)=0$, $(\cos x+\sin x)(2 \cos x(\cos x-\sin x)-1)=0, \quad(\cos x+\sin x) \times$ $\times\left(2 \cos ^{2} x-2 \sin x \cos x... | x_{1}=\frac{\pi}{4}(4k-1),x_{2}=\frac{\pi}{8}(4n+1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,033 |
8.148. $\operatorname{tg} 2 x \sin 2 x-3 \sqrt{3} \operatorname{ctg} 2 x \cos 2 x=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 2 x \neq 0, \\ \sin 2 x \neq 0 .\end{array}\right.$
## We have
$$
\begin{aligned}
& \frac{\sin 2 x \sin 2 x}{\cos 2 x}-\frac{3 \sqrt{3} \cos 2 x \cos 2 x}{\sin 2 x}=0 \Rightarrow \sin ^{3} 2 x-3 \sqrt{3} \cos ^{3} 2 x=0 \\
& \operatorname{tg}^{3} 2 x=3 ... | \frac{\pi}{6}(3n+1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,034 |
8.149. $\cos x - \cos 3x = \sin 2x$. | Solution.
From the condition
$$
\begin{aligned}
& 2 \sin \frac{x+3 x}{2} \sin \frac{3 x-x}{2}-\sin 2 x=0, \quad 2 \sin 2 x \sin x-\sin 2 x=0 \\
& \sin 2 x(2 \sin x-1)=0
\end{aligned}
$$
## Hence
1) $\sin 2 x=0, 2 x=\pi k, \quad x_{1}=\frac{\pi \cdot k}{2}, k \in Z$;
2) $2 \sin x-1=0, \sin x=\frac{1}{2}, \quad x_{2}... | x_{1}=\frac{\pik}{2};x_{2}=(-1)^{n}\frac{\pi}{6}+\pin,k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,035 |
8.152. $\sin ^{2} 3 x=3 \cos ^{2} 3 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.152. $\sin ^{2} 3 x=3 \cos ^{2} 3 x$. | Solution.
Dividing this equation by $\cos ^{2} 3 x \neq 0$, we get
$\operatorname{tg}^{2} 3 x=3, \operatorname{tg} 3 x= \pm \sqrt{3}, 3 x= \pm \frac{\pi}{3}+\pi k, x= \pm \frac{\pi}{9}+\frac{\pi k}{3}=\frac{\pi}{9}(3 k \pm 1), k \in Z$
Answer: $\quad x=\frac{\pi}{9}(3 k \pm 1), k \in Z$ | \frac{\pi}{9}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,037 |
8.154. $\sin 6 x+\sin 2 x=\frac{1}{2} \operatorname{tg} 2 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.154. $\sin 6 x+\sin 2 x=\frac{1}{2} \tan 2 x$. | ## Solution.
Domain of definition: $\cos 2 x \neq 0$.
Using the formula $\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}$, we get
$2 \sin \frac{6 x+2 x}{2} \cos \frac{6 x-2 x}{2}-\frac{\sin 2 x}{2 \cos 2 x}=0, 2 \sin 4 x \cos 2 x-\frac{\sin 2 x}{2 \cos 2 x}=0$, $4 \sin 2 x \cos 2 x \... | x_{1}=\frac{\pin}{2};x_{2}=\frac{\pi}{6}(6k\1),n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,038 |
8.155. $\frac{2 \cos (\pi+x)-5 \cos \left(\frac{3}{2} \pi-x\right)}{\cos \left(\frac{3}{2} \pi+x\right)-\cos (\pi-x)}=\frac{3}{2}$. | ## Solution.
Domain of definition: $\sin x+\cos x \neq 0$.
Using reduction formulas
$$
\begin{aligned}
& \frac{-2 \cos x+5 \sin x}{\sin x+\cos x}-\frac{3}{2}=0 \Rightarrow 7 \sin x-7 \cos x=0 \Leftrightarrow \operatorname{tg} x=1 \\
& x=\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k+1), \quad k \in Z
\end{aligned}
$$
Answer... | \frac{\pi}{4}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,039 |
8.156. $(\sin 2 x+\sqrt{3} \cos 2 x)^{2}=2-2 \cos \left(\frac{2}{3} \pi-x\right)$. | Solution.
We have
$$
\begin{aligned}
& 4\left(\frac{1}{2} \sin 2 x+\frac{\sqrt{3}}{2} \cos 2 x\right)^{2}=2-2 \cos \left(\frac{2}{3} \pi-x\right) \\
& 2\left(\sin \frac{\pi}{6} \sin 2 x+\cos \frac{\pi}{6} \cos 2 x\right)^{2}=1-\cos \left(\frac{2}{3} \pi-x\right) \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{2}\left(\... | x_{1}=\frac{2}{5}\pin,x_{2}=\frac{2\pi}{9}(3k+1),n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,040 |
8.157. $\operatorname{ctg} x+\operatorname{tg} 2 x+1=4 \cos ^{2} x+\frac{\sin 3 x}{\sin x}-2 \cos 2 x$.
8.157. $\cot x + \tan 2x + 1 = 4 \cos^2 x + \frac{\sin 3x}{\sin x} - 2 \cos 2x$. | ## Solution.
$$
\text { domain: }\left\{\begin{array}{l}
\cos 2 x \neq 0 \\
\sin x \neq 0
\end{array}\right.
$$
Using the formula $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we can rewrite the equation as
$$
\frac{\cos x}{\sin x}+\frac{\sin 2 x}{\cos 2 x}+1=4 \cos ^{2} x+\frac{3 \sin x-4 \sin ^{3} x}{\sin x}-2... | x_{1}=\frac{\pi}{2}(2k+1),x_{2}=\frac{\pi}{8}(4n+1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,041 |
8.158. $\operatorname{tg} x \operatorname{tg} 20^{\circ}+\operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}+\operatorname{tg} 40^{\circ} \operatorname{tg} x=1$. | ## Solution.
Domain of definition: $\cos x \neq 0$.
By the condition
$\frac{\sin x \sin 20^{\circ}}{\cos x \cos 20^{\circ}}+\frac{\sin 20^{\circ} \sin 40^{\circ}}{\cos 20^{\circ} \cos 40^{\circ}}+\frac{\sin 40^{\circ} \sin x}{\cos 40^{\circ} \cos x}-1=0$,
$\left(\sin x \sin 20^{\circ} \cos 40^{\circ}+\sin x \cos 20... | 30+180k,\quadk\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,042 |
8.159. $2 \cos ^{2} \frac{x}{2}-1=\sin 3 x$.
8.159. $2 \cos ^{2} \frac{x}{2}-1=\sin 3 x$.
The above text is already in a mathematical equation format, which is the same in both Chinese and English. Therefore, the translation is identical to the original text. | ## Solution.
We have
$$
\begin{aligned}
& \cos x - \cos \left(\frac{\pi}{2} - 3 x\right) = 0 \Leftrightarrow \sin \left(\frac{\pi}{4} - x\right) \sin \left(\frac{\pi}{4} - 2 x\right) = 0 \\
& \sin \left(x - \frac{\pi}{4}\right) \sin \left(2 x - \frac{\pi}{4}\right) = 0
\end{aligned}
$$
From this:
1) $\sin \left(x -... | x_{1}=\frac{\pi}{4}(4k+1),\quadx_{2}=\frac{\pi}{8}(4n+1),\quadk,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,043 |
8.161. $3 \cos ^{2} x=\sin ^{2} x+\sin 2 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.161. $3 \cos ^{2} x=\sin ^{2} x+\sin 2 x$. | Solution.
From the condition $\sin ^{2} x+2 \sin x \cos x-3 \cos ^{2} x=0$. Dividing this equation by $\cos ^{2} x \neq 0$, we get $\operatorname{tg}^{2} x+2 \operatorname{tg} x-3=0$. Solving the equation as a quadratic equation in terms of $\operatorname{tg} x$, we find $(\operatorname{tg} x)_{1}=-3, x_{1}=-\operator... | x_{1}=-\operatorname{arctg}3+\pik,x_{2}=\frac{\pi}{4}(4n+1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,044 |
8.162. $2(1-\cos 2 x)=\sqrt{3} \tan x$. | Solution.
Domain of definition: $\cos x \neq 0$.
From the condition
$$
2(1-\cos 2 x)-\frac{\sqrt{3}(1-\cos 2 x)}{\sin 2 x}=0,(1-\cos 2 x) \cdot\left(2-\frac{\sqrt{3}}{\sin 2 x}\right)=0
$$
## Hence
1) $1-\cos 2 x=0, \cos 2 x=1,2 x=2 \pi k, x_{1}=\pi k, k \in Z$
2) $2-\frac{\sqrt{3}}{\sin 2 x}=0, \sin 2 x=\frac{\sq... | x_{1}=\pik,x_{2}=(-1)^{n}\frac{\pi}{6}+\frac{\pin}{2},k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,045 |
8.163. $a \cos ^{2} \frac{x}{2}-(a+2 b) \sin ^{2} \frac{x}{2}=a \cos x-b \sin x ; b \neq 0$. | ## Solution.
Using the formulas for reducing the power
$$
\begin{aligned}
& \frac{a}{2}(1+\cos x)-\frac{a+2 b}{2} \cdot(1-\cos x)-a \cos x+b \sin x=0, a+a \cos x-a-2 b+ \\
& +(a+2 b) \cos x-2 a \cos x+2 b \sin x=0 \Leftrightarrow 2 b \cos x+2 b \sin x-2 b=0 \\
& b(\cos x+\sin x-1)=0
\end{aligned}
$$
Since $b \neq 0$... | x_{1}=2\pin;x_{2}=\frac{\pi}{2}(4k+1)\quadn,k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 51,046 |
8.164. $\sin 5 x=\cos 4 x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.164. $\sin 5 x=\cos 4 x$. | Solution.
Using the reduction formulas
$$
\begin{aligned}
& \sin 5 x-\sin \left(\frac{\pi}{2}-4 x\right)=0 \Leftrightarrow 2 \sin \frac{5 x-\frac{\pi}{2}+4 x}{2} \cos \frac{5 x+\frac{\pi}{2}-4 x}{2}=0, \\
& \sin \left(\frac{9 x}{2}-\frac{\pi}{4}\right) \cos \left(\frac{x}{2}+\frac{\pi}{4}\right)=0 .
\end{aligned}
$$
... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,047 | |
8.165. $2 \tan x - 2 \cot x = 3$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0 \\ \sin x \neq 0\end{array}\right.$
We have
$2\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right)=3, \frac{\sin ^{2} x-\cos ^{2} x}{\sin x \cos x}=\frac{3}{2} \Leftrightarrow \frac{\cos 2 x}{\sin 2 x}=-\frac{3}{4}$, $\operatorname{ctg} 2 x=-\... | -\frac{1}{2}\operatorname{arcctg}\frac{3}{4}+\frac{\pin}{2},n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,048 |
8.166. $25 \sin ^{2} x+100 \cos x=89$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.166. $25 \sin ^{2} x+100 \cos x=89$. | ## Solution.
From the condition
$$
25\left(1-\cos ^{2} x\right)+100 \cos x-89=0,25 \cos ^{2} x-100 \cos x+64=0
$$
Solving the equation as a quadratic in terms of $\cos x$, we get $\cos x=\frac{16}{5}, \varnothing$, or $\cos x=\frac{4}{5}, x= \pm \arccos \frac{4}{5}+2 \pi k, k \in Z$.
Answer: $\quad x= \pm \arccos \... | \\arccos\frac{4}{5}+2\pik,\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,049 |
8.169. $\sin x + \sin 3x = 4 \cos^3 x$. | ## Solution.
From the condition
$$
\begin{aligned}
& 2 \sin \frac{x+3 x}{2} \cos \frac{x-3 x}{2}-4 \cos ^{3} x=0, \quad 2 \sin 2 x \cos x-4 \cos ^{3} x=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos x\left(2 \sin x \cos x-2 \cos ^{2} x\right)=0, \quad 4 \cos ^{2} x(\sin x-\cos x)=0
\end{aligned}
$$
From this:
1) $\c... | x_{1}=\frac{\pi}{2}(2k+1),x_{2}=\frac{\pi}{4}(4n+1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,050 |
8.170. $\cos 2 x+3 \sin x=2$.
8.170. $\cos 2 x+3 \sin x=2$.
The translation is the same as the original text because it is a mathematical equation, which is universally understood and does not change when translated into another language. | ## Solution.
We have $1-2 \sin ^{2} x+3 \sin x-2=0, 2 \sin ^{2} x-3 \sin x+1=0$. Solving this equation as a quadratic in terms of $\sin x$, we get $(\sin x)_{1}=\frac{1}{2}$,
$$
x_{1}=(-1)^{k} \frac{\pi}{6}+\pi k, k \in Z ;(\sin x)_{2}=1, x_{2}=\frac{\pi}{2}+2 \pi n=\frac{\pi}{2}(4 n+1) ; n \in Z
$$
Answer: $\quad x... | x_{1}=(-1)^{k}\frac{\pi}{6}+\pik;x_{2}=\frac{\pi}{2}(4n+1)\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,051 |
8.174. $\operatorname{tg}^{2} 3 x-2 \sin ^{2} 3 x=0$.
8.174. $\tan^{2} 3 x-2 \sin ^{2} 3 x=0$. | ## Solution.
Domain of definition: $\cos 3 x \neq 0$.
We have $\frac{\sin ^{2} 3 x}{\cos ^{2} 3 x}-2 \sin ^{2} 3 x=0, \sin ^{2} 3 x \cdot\left(\frac{1}{\cos ^{2} 3 x}-2\right)=0$.
From this:
1) $\sin 3 x=0, \quad 3 x=\pi k, \quad x_{1}=\frac{\pi k}{3}, k \in Z$;
2) $\frac{1}{\cos ^{2} 3 x}-2=0, \cos 3 x= \pm \frac{... | x_{1}=\frac{\pik}{3};x_{2}=\frac{\pi}{12}(2n+1)\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,052 |
9.001. Show that for all positive numbers $a$ and $b$ the inequality $\sqrt{a}+\sqrt{b}>\sqrt{a+b}$ holds. | ## Solution.
By squaring both sides of the given inequality, we have the equivalent inequality $a+2 \sqrt{a b}+b>a+b \Leftrightarrow 2 \sqrt{a b}>0 \Leftrightarrow \sqrt{a b}>0, a b>0$. Since $a>0$ and $b>0$, the last inequality is obvious, and thus the validity of the original inequality, which is equivalent to it, i... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 51,053 |
9.002. Prove that if $a>0$ and $b>0$, then $\frac{2 \sqrt{a b}}{\sqrt{a}+\sqrt{b}} \leq \sqrt[4]{a b}$. | ## Solution.
Since $\sqrt{a}+\sqrt{b}>0$, we get
$$
2 \sqrt{a b} \leq(\sqrt{a}+\sqrt{b}) \sqrt[4]{a b} \Leftrightarrow 2 \sqrt{a b}-(\sqrt{a}+\sqrt{b}) \sqrt[4]{a b} \leq 0
$$
considering that $-\sqrt[4]{a b}<0$, we find $\sqrt{a}-2 \sqrt[4]{a b}+\sqrt{b} \geq 0 \Leftrightarrow(\sqrt[4]{a}-\sqrt[4]{b})^{2} \geq 0$. ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 51,054 |
9.004. Prove that if $a \neq 2$, then $\frac{1}{a^{2}-4 a+4}>\frac{2}{a^{3}-8}$. | Solution.
Rewrite the given inequality as
$$
\begin{aligned}
& \frac{1}{(a-2)^{2}}-\frac{2}{(a-2)\left(a^{2}+2 a+4\right)}>0 \Leftrightarrow \frac{a^{2}+2 a+4-2 a+4}{(a-2)^{2}\left(a^{2}+2 a+4\right)}>0 \Leftrightarrow \\
& \Leftrightarrow \frac{a^{2}+8}{(a-2)^{2}\left(a^{2}+2 a+4\right)}>0 .
\end{aligned}
$$
Since ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 51,055 |
9.010. Find the positive integer values of $x$ that satisfy the inequality $\frac{5 x+1}{x-1}>2 x+2$. | Solution.
We have
$$
\begin{aligned}
& \frac{5 x+1}{x-1}-2 x-2>0 \Leftrightarrow \frac{5 x+1-2(x+1)(x-1)}{x-1}>0 \Leftrightarrow \frac{-2 x^{2}+5 x+3}{x-1}>0 \Leftrightarrow \\
& \Leftrightarrow\left(2 x^{2}-5 x-3\right)(x-1)<0 \Leftrightarrow 2\left(x+\frac{1}{2}\right)(x-3)(x-1)<0
\end{aligned}
$$
Using the number... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,059 |
9.012. Find the natural values of $x$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
\log _{\sqrt{2}}(x-1)<4 \\
\frac{x}{x-3}+\frac{x-5}{x}<\frac{2 x}{3-x}
\end{array}\right.
$$ | Solution.
From the condition
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 0 0 \text { for } x \in R, \\
x(x-3)<0 .
\end{array}\right.
\end{aligned}
$$
Using the number line, we find the solution to the system $x=2$.
<1
\end{array}\right.
$$ | Solution.
Considering the domain of definition, we solve the second inequality of the system:
$$
\left\{\begin{array} { l }
{ \frac { x + 8 } { x + 2 } - 2 > 0 , } \\
{ 0 0 } \\
{ 1 < x < 1 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x-4)(x+2)<0 \\
1<x<11
\end{array}\right.\right.\right.
$$
Using a nu... | x_{1}=2;x_{2}=3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,063 |
9.018. $y=\sqrt{\log _{0.3} \frac{x-1}{x+5}}$. | ## Solution.
$D(y):\left\{\begin{array}{l}\frac{x-1}{x+5}>0, \\ \log _{0.3} \frac{x-1}{x+5} \geq 0,\end{array}\left\{\begin{array}{l}\frac{x-1}{x+5}>0, \\ \frac{x-1}{x+5} \leq 1, \\ x+5 \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}(x-1)(x+5)>0, \\ x+5>0 .\end{array}\right.\right.\right.$
Using the interval... | x\in(1;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,065 |
9.020. $y=\sqrt{5-x-\frac{6}{x}}$.
The above text is translated into English as follows, retaining the original text's line breaks and format:
9.020. $y=\sqrt{5-x-\frac{6}{x}}$. | ## Solution.
$$
D(y): 5-x-\frac{6}{x} \geq 0,\left\{\begin{array}{l}
x\left(x^{2}-5 x+6\right) \leq 0, \\
x \neq 0
\end{array},\left\{\begin{array}{l}
x(x-2)(x-3) \leq 0 \\
x \neq 0
\end{array}\right.\right.
$$
Using the interval method, we find $x<0$ or $2 \leq x \leq 3$.
\cup[2;3] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,067 |
9.021. $y=\sqrt{-\frac{\log _{0.3}(x-1)}{\sqrt{-x^{2}+2 x+8}}}$.
9.021. $y=\sqrt{-\frac{\log _{0.3}(x-1)}{\sqrt{-x^{2}+2 x+8}}}$. | ## Solution.
$$
\begin{aligned}
& -\frac{\log _{0.3}(x-1)}{\sqrt{-x^{2}+2 x+8}} \geq 0, \frac{\log _{0.3}(x-1)}{\sqrt{-x^{2}+2 x+8}} \leq 0,\left\{\begin{array}{l}
\log _{0.3}(x-1) \leq 0, \\
-x^{2}+2 x+8>0
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l }
{ x - 1 \geq 1 , } \\
{ x ^ {... | x\in[2;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,068 |
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