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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
8.288. $(\operatorname{ctg} z-1)(1+\sin 2 z)=1+\operatorname{ctg} z$.
8.288. $(\cot z-1)(1+\sin 2z)=1+\cot z$. | ## Solution.
Domain of definition: $\sin z \neq 0$.
Rewrite the equation as
$$
\begin{aligned}
& \left(\frac{\cos z}{\sin z}-1\right) \cdot\left(\cos ^{2} z+2 \sin z \cos z+\sin ^{2} z\right)-\left(1+\frac{\cos z}{\sin z}\right)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{(\cos z-\sin z)(\cos z+\sin z)^{2}}{\sin z}... | \frac{\pi}{4}(4n-1),n\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,577 |
8.290. $\sin ^{4} 3 t+\sin ^{4}\left(\frac{\pi}{4}+3 t\right)=\frac{1}{4}$ | Solution.
Write the equation in the form
$\left(\sin ^{2} 3 t\right)^{2}+\left(\sin ^{2}\left(\frac{\pi}{4}+3 t\right)\right)^{2}-\frac{1}{4}=0 \Leftrightarrow$
$\Leftrightarrow\left(\frac{1}{2} \cdot(1-\cos 6 t)\right)^{2}+\left(\frac{1}{2} \cdot\left(1-\cos \left(\frac{\pi}{2}+6 t\right)\right)\right)^{2}-\frac{1}... | t_{1}=\frac{\pik}{3};t_{2}=\frac{\pi}{12}(4n-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,579 |
8.291. $\cos 10 x+2 \cos ^{2} 4 x+6 \cos 3 x \cos x=\cos x+8 \cos x \cos ^{3} 3 x$. | ## Solution.
We have:
$\cos 10 x+(1+\cos 8 x)+3(2 \cos 3 x \cos x)=\cos x+2(2 \cos x \cos 3 x) \cos ^{2} 3 x \Leftrightarrow$ $\Leftrightarrow \cos 10 x+1+\cos 8 x+3(\cos 2 x+\cos 4 x)=$
$=\cos x+2(\cos 2 x+\cos 4 x) \cdot(1+\cos 6 x)$
$\cos 10 x+1+\cos 8 x+3 \cos 2 x+3 \cos 4 x=\cos x+2 \cos 2 x+$
$+2 \cos 4 x+2 ... | 2\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,580 |
8.292. $1+\sin \frac{t}{2} \sin t-\cos \frac{t}{2} \sin ^{2} t=2 \cos ^{2}\left(\frac{\pi}{4}-\frac{t}{2}\right)$.
8.292. $1+\sin \frac{t}{2} \sin t-\cos \frac{t}{2} \sin ^{2} t=2 \cos ^{2}\left(\frac{\pi}{4}-\frac{t}{2}\right)$. | ## Solution.
Let's write the equation as
$$
\begin{aligned}
& 1+\sin \frac{t}{2} \sin t-\cos \frac{t}{2} \sin ^{2} t=1+\cos \left(\frac{\pi}{2}-t\right) \Leftrightarrow \\
& \Leftrightarrow \sin \frac{t}{2} \sin t-\cos \frac{t}{2} \sin ^{2} t-\sin t=0, \sin t \cdot\left(\sin \frac{t}{2}-\cos \frac{t}{2} \sin t-1\righ... | \pik,k\in\mathbb{Z} | Algebra | proof | Yes | Yes | olympiads | false | 48,581 |
8.293. $\frac{4 \sin \left(\frac{\pi}{6}+x\right) \sin \left(\frac{5 \pi}{6}+x\right)}{\cos ^{2} x}+2 \tan x=0$. | ## Solution.
Domain of definition: $\cos x \neq 0$.
Write the equation in the form
$$
\begin{aligned}
& \frac{2\left(\cos \left(\frac{\pi}{6}+x-\frac{5 \pi}{6}-x\right)-\cos \left(\frac{\pi}{6}+x+\frac{5 \pi}{6}+x\right)\right)}{\cos ^{2} x}+\frac{2 \sin x}{\cos x}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2\left... | x_{1}=-\operatorname{arctg}\frac{1}{3}+\pik;x_{2}=\frac{\pi}{4}(4n+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,582 |
8.294. $\frac{4 \cos ^{2} t-1}{\sin t}=\operatorname{ctg} t(1+2 \cos 2 t)$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.294. $\frac{4 \cos ^{2} t-1}{\sin t}=\operatorname{ctg} t(1+2 \cos 2 t)$.
The above text in English, k... | ## Solution.
Domain of definition: $\sin t \neq 0$.
Write the equation in the form
$\frac{4 \cos ^{2} t-1}{\sin t}=\frac{\cos t\left(1+2\left(2 \cos ^{2} t-1\right)\right)}{\sin t} \Leftrightarrow$
$\Leftrightarrow 4 \cos ^{2} t-1=\cos t\left(4 \cos ^{2} t-1\right),\left(4 \cos ^{2} t-1\right)-\cos t\left(4 \cos ^{... | \frac{\pi}{3}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,583 |
8.295. $(\sin x+\cos x)^{4}=2\left(1+\sin ^{2} x\right)-(\sin x-\cos x)^{4}$. | ## Solution.
Let's rewrite the equation as
$\left((\sin x+\cos x)^{2}\right)^{2}+\left((\sin x-\cos x)^{2}\right)^{2}=2\left(1+\sin ^{2} x\right) \Leftrightarrow$
$\Leftrightarrow\left(\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x\right)^{2}+\left(\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x\right)^{2}=$ $=2\left(1+\sin ^{2} x... | x_{1}=\pik,x_{2}=\frac{\pi}{3}(3\1) | Algebra | proof | Yes | Yes | olympiads | false | 48,584 |
8.296. $\cos ^{-4} z=64 \cos ^{2} 2 z$. | Solution.
Domain of definition: $\cos z \neq 0$.
From the condition we have
$\frac{1}{\cos ^{4} z}=(8 \cos 2 z)^{2} \Leftrightarrow\left[\begin{array}{l}\frac{1}{\cos ^{2} z}=8 \cos 2 z, \cos 2 z>0, \\ \frac{1}{\cos ^{2} z}=-8 \cos 2 z, \cos 2 z<0,\end{array} \Leftrightarrow\right.$
$\Leftrightarrow\left[\begin{arr... | z_{1}=\\arccos\frac{\sqrt{1+\sqrt{2}}}{2}+\pik,k\inZ;z_{2}=\frac{\pi}{3}(3n\1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,585 |
8.297. $4 \sin 5 x \cos 5 x\left(\cos ^{4} x-\sin ^{4} x\right)=\sin 4 x$.
8.297. $4 \sin 5 x \cos 5 x\left(\cos ^{4} x-\sin ^{4} x\right)=\sin 4 x$. | ## Solution.
Let's write this equation in the form
$2 \sin 10 x\left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)-2 \sin 2 x \cos 2 x=0 \Leftrightarrow$
$\Leftrightarrow 2 \sin 10 x \cos 2 x-2 \sin 2 x \cos 2 x=0, 2 \cos 2 x(\sin 10 x-\sin 2 x)=0$.
From this,
1) $\cos 2 x=0, 2 x=\frac{\pi}{2}... | x_{1}=\frac{\pik}{4};x_{2}=\frac{\pi}{12}(2n+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,586 |
8.299. $\frac{\tan x + \cot x}{\cot x - \tan x} = 6 \cos 2x + 4 \sin 2x$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\operatorname{ctg} x-\operatorname{tg} x \neq 0, \\ \cos x \neq 0, \\ \sin x \neq 0 .\end{array}\right.$
## Write the equation in the form
;z_{2}=\frac{\pi}{4}(4n-1),k,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,590 |
8.303. $\left(\cos ^{-6} z-\operatorname{tg}^{6} z-\frac{7}{3}\right) \cdot(\sin z+\cos z+2)=0$.
Translate the above text into English, keeping the original text's line breaks and format:
8.303. $\left(\cos ^{-6} z-\tan^{6} z-\frac{7}{3}\right) \cdot(\sin z+\cos z+2)=0$. | ## Solution.
Domain of definition: $\cos z \neq 0$.
From the condition we have
$$
\begin{aligned}
& {\left[\frac{1}{\cos ^{6} z}-\frac{\sin ^{6} z}{\cos ^{6} z}-\frac{7}{3}=0, \Leftrightarrow 3-3 \sin ^{6} z-7 \cos ^{6} z=0 \Leftrightarrow\right.} \\
& \Leftrightarrow 3-3\left(\sin ^{2} z\right)^{3}-7\left(\cos ^{2}... | \frac{\pi}{6}(6n\1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,592 |
8.304. $\operatorname{tg} 2 x-\operatorname{ctg} 3 x+\operatorname{ctg} 5 x=0$.
8.304. $\tan 2x - \cot 3x + \cot 5x = 0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 2 x \neq 0, \\ \sin 3 x \neq 0, \\ \sin 5 x \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& \left(\frac{\sin 2 x}{\cos 2 x}-\frac{\cos 3 x}{\sin 3 x}\right)+\frac{\cos 5 x}{\sin 5 x}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{... | x_{1}=\frac{\pi}{10}(2k+1);x_{2}=\frac{\pi}{6}(2+1),k,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,593 |
8.308. $\cos 3z - \cos^3 z + \frac{3}{4} \sin 2z = 0$. | Solution.
Since $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$, we obtain
$$
\begin{aligned}
& 4\left(4 \cos ^{3} z-3 \cos z\right)-4 \cos ^{3} z+6 \sin z \cos z=0 \\
& 12 \cos ^{3} z-12 \cos z+6 \sin z \cos z=0,6 \cos z\left(2 \cos ^{2} z-2+\sin z\right)=0
\end{aligned}
$$
$6 \cos z\left(2\left(1-\sin ^{2} z\righ... | z_{1}=\frac{\pi}{2}(2k+1);z_{2}=\pin;z_{3}=(-1)^{k}\frac{\pi}{6}+\pi, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,596 |
8.309. $\operatorname{tg}\left(t^{2}-t\right) \operatorname{ctg} 2=1$.
8.309. $\tan\left(t^{2}-t\right) \cot 2=1$. | ## Solution.
Domain of definition: $\cos \left(t^{2}-t\right) \neq 0$.
From the condition we have
$\operatorname{tg}\left(t^{2}-t\right)=\operatorname{tg} 2, \operatorname{tg}\left(t^{2}-t\right)-\operatorname{tg} 2=0 \Leftrightarrow \frac{\sin \left(t^{2}-t-2\right)}{\cos \left(t^{2}-t\right) \cos 2}=0 \Leftrightar... | t_{1,2}=\frac{1\\sqrt{9+4\pik}}{2},wherek=0;1;2;\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,597 |
8.310. $\sin ^{3} x(1-\operatorname{ctg} x)+\cos ^{3} x(1-\operatorname{tg} x)=1.5 \cos 2 x$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0 \\ \cos x \neq 0 .\end{array}\right.$
## Write the equation in the form
$$
\begin{aligned}
& \sin ^{3} x\left(1-\frac{\cos x}{\sin x}\right)+\cos ^{3} x\left(1-\frac{\sin x}{\cos x}\right)-1.5 \cos 2 x=0 \Leftrightarrow \\
& \Leftrightarrow \sin ^... | \frac{\pi}{4}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,598 |
8.311. $\frac{\cos ^{2}\left(\frac{\pi}{2}-2 t\right)}{1+\cos 2 t}=\cos ^{-2} 2 t-1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 2 t \neq 0, \\ \cos 2 t \neq-1 .\end{array}\right.$
From the condition we have
$$
\begin{aligned}
& \frac{\sin ^{2} 2 t}{1+\cos 2 t}-\frac{1}{\cos ^{2} 2 t}+1=0, \frac{1-\cos ^{2} 2 t}{1+\cos 2 t}-\frac{1-\cos ^{2} 2 t}{\cos ^{2} 2 t}=0 \\
& \frac{(1-\c... | t_{1}=\pik;t_{2}=\\frac{1}{2}\arccos\frac{1-\sqrt{5}}{2}+\pin,k,n\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,599 |
8.312. $4 \cos x \cos 2 x \cos 3 x=\cos 6 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.312. $4 \cos x \cos 2 x \cos 3 x=\cos 6 x$. | Solution.
Multiply both sides of the equation by $\sin x \neq 0$. We get
$2(2 \sin x \cos x) \cos 2 x \cos 3 x=\sin x \cos 6 x \Leftrightarrow$
$\Leftrightarrow(2 \sin 2 x \cos 2 x) \cos 3 x=\sin x \cos 6 x \Leftrightarrow$
$\Leftrightarrow \sin 4 x \cos 3 x-\sin x \cos 6 x=0 \Leftrightarrow$
$\Leftrightarrow \fra... | x_{1}=\frac{\pi}{3}(3\1);x_{2}=\frac{\pi}{4}(2n+1) | Algebra | proof | Yes | Yes | olympiads | false | 48,600 |
8.313. $1-\cos x=\sqrt{1-\sqrt{4 \cos ^{2} x-7 \cos ^{4} x}}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}4 \cos ^{2} x-7 \cos ^{4} x \geq 0, \\ 1-\sqrt{4 \cos ^{2} x-7 \cos ^{4} x} \geq 0 .\end{array}\right.$
Squaring both sides of the equation, we get
$$
\left\{\begin{array}{l}
1-2 \cos x+\cos ^{2} x=1-\sqrt{4 \cos ^{2} x-7 \cos ^{4} x} \\
1-\cos x \geq 0
\end... | x_{1}=\frac{\pi}{2}(2k+1);x_{2}=\frac{\pi}{3}(6n\1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,601 |
8.314. $\frac{2 \sin x-\sin 2 x}{2 \sin x+\sin 2 x}+\operatorname{ctg}^{2} \frac{x}{2}=\frac{10}{3}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin \frac{x}{2} \neq 0, \\ 2 \sin x+\sin 2 x \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& \frac{2 \sin x-2 \sin x \cos x}{2 \sin x+2 \sin x \cos x}+\frac{1+\cos x}{1-\cos x}-\frac{10}{3}=0 \\
& \frac{2 \sin x(1-\cos x)}{2 \sin x(... | \frac{\pi}{3}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,602 |
8.315. $4\left(\sin t \cos ^{5} t+\cos t \sin ^{5} t\right)+\sin ^{3} 2 t=1$. | ## Solution.
We have
$4 \sin t \cos t\left(\cos ^{4} t+\sin ^{4} t\right)+\sin ^{3} 2 t-1=0$,
$2 \sin 2 t\left(\left(\cos ^{2} t+\sin ^{2} t\right)^{2}-2 \cos ^{2} t \sin ^{2} t\right)+\sin ^{3} 2 t-1=0$,
$2 \sin 2 t-\sin 2 t\left(4 \cos ^{2} t \sin ^{2} t\right)+\sin ^{3} 2 t-1=0$
$2 \sin 2 t-\sin ^{3} 2 t+\sin ^... | (-1)^{k}\frac{\pi}{12}+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,603 |
8.316. $\sin ^{4} x-\sin ^{2} x+4(\sin x+1)=0$.
8.316. $\sin ^{4} x-\sin ^{2} x+4(\sin x+1)=0$. | ## Solution.
From the condition we have
$$
\begin{aligned}
& \sin ^{2} x\left(\sin ^{2} x-1\right)+4(\sin x+1)=0 \\
& \sin ^{2} x(\sin x-1)(\sin x+1)+4(\sin x+1)=0 \\
& (\sin x+1)\left(\sin ^{3} x-\sin ^{2} x+4\right)=0
\end{aligned}
$$
From this,
1) $\sin x=-1, x=-\frac{\pi}{2}+\pi k=\frac{\pi}{2}(4 k-1), k \in Z$... | \frac{\pi}{2}(4k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,604 |
8.317. $\frac{\sin ^{2} t-\tan^{2} t}{\cos ^{2} t-\cot^{2} t}+2 \tan^{3} t+1=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \sin t \neq 0, \\ \sin t \neq \pm 1 .\end{array}\right.$
Rewrite the equation as
$\frac{\sin ^{2} t-\frac{\sin ^{2} t}{\cos ^{2} t}}{\cos ^{2} t-\frac{\cos ^{2} t}{\sin ^{2} t}}+2 \operatorname{tg}^{3} t+1=0 \Leftrightarrow$
$\Leftrightarr... | \frac{\pi}{4}(4k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,605 |
8.318. $\frac{\tan t}{\cos ^{2} 5 t}-\frac{\tan 5 t}{\cos ^{2} t}=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \cos 5 t \neq 0 .\end{array}\right.$
## Write the equation in the form
$\frac{\sin t}{\cos t \cos ^{2} 5 t}-\frac{\sin 5 t}{\cos 5 t \cos ^{2} t}=0 \Leftrightarrow \sin t \cos t-\sin 5 t \cos 5 t=0$.
Using the formula $\sin \alpha \cos \al... | t_{1}=\frac{\pi}{12}(2k+1);t_{2}=\pin | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,606 |
8.319. $\frac{1+\sin 2 x+\cos 2 x}{1+\sin 2 x-\cos 2 x}+\sin x\left(1+\operatorname{tg} x \operatorname{tg} \frac{x}{2}\right)=4$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos \frac{x}{2} \neq 0, \\ 1+\sin 2 x-\cos 2 x \neq 0 .\end{array}\right.$
From the condition we have
$\frac{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x+\cos ^{2} x-\sin ^{2} x}{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x-\cos ^{2} x+\sin ^{2} x}+\si... | (-1)^{k}\frac{\pi}{12}+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,607 |
8.320. $\sin 2z - \sin 6z + 2 = 0$. | Solution.
Write the equation as $\sin 2 z-\sin 3(2 z)+2=0$. Using the formula $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we have
$$
\begin{aligned}
& \sin 2 z-3 \sin 2 z+4 \sin ^{3} 2 z+2=0,4 \sin ^{3} 2 z-2 \sin 2 z+2=0 \\
& 2 \sin ^{3} 2 z-\sin 2 z+1=0,2 \sin ^{3} 2 z+2-\sin 2 z-1=0
\end{aligned}
$$
$2(\sin... | \frac{\pi}{4}(4k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,608 |
8.321. $\sin ^{2}\left(t+45^{\circ}\right)-\sin ^{2}\left(t-30^{\circ}\right)-\sin 15^{\circ} \cos \left(2 t+15^{\circ}\right)=0.5 \sin 6 t$. | ## Solution.
From the condition, we have
$$
\begin{aligned}
& 1-\cos \left(2 t+90^{\circ}\right)-1+\cos \left(2 t-60^{\circ}\right)-2 \sin 15^{\circ} \cos \left(2 t+15^{\circ}\right)-\sin 6 t=0 \Leftrightarrow \\
& \Leftrightarrow \sin 2 t+\cos \left(2 t-60^{\circ}\right)+\sin 2 t-\sin \left(2 t+30^{\circ}\right)-\si... | t_{1}=90k,t_{2}=\15+90,k,\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,609 |
8.322. $3 \tan 3 x-4 \tan 2 x=\tan^{2} 2 x \tan 3 x$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 3 x \neq 0, \\ \cos 2 x \neq 0 .\end{array}\right.$
## Write the equation in the form
$$
\begin{aligned}
& 3 \operatorname{tg} 3 x-3 \operatorname{tg} 2 x=\operatorname{tg}^{2} 2 x \operatorname{tg} 3 x+\operatorname{tg} 2 x \\
& 3 \cdot\left(\frac{\sin... | \pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,610 |
8.323. $\frac{5 \sin x-5 \tan x}{\sin x+\tan x}+4(1-\cos x)=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0, \\ \cos x \neq-1 .\end{array}\right.$
From the condition we have
$$
\begin{aligned}
& \frac{5 \sin x-\frac{5 \sin x}{\cos x}}{\sin x+\frac{\sin x}{\cos x}}+4(1-\cos x)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{5 \sin x(\co... | \\arccos\frac{1}{4}+2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,611 |
8.324. $4 \cos x=\sqrt{3} \operatorname{ctg} x+1$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.324. $4 \cos x=\sqrt{3} \cot x+1$. | ## Solution.
Domain of definition: $\sin x \neq 0$.
Let's write the equation in the form
$$
\begin{aligned}
& 4 \cos x - \frac{\sqrt{3} \cos x}{\sin x} - 1 = 0 \Leftrightarrow 2 \sin x \cos x - \left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right) = 0, \\
& \sin 2 x - \left(\sin \frac{\pi}{3} \cos x + \cos \fr... | x_{1}=\frac{2}{9}\pi(3k+1),x_{2}=\frac{\pi}{3}(6n+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,612 |
8.326. $(\cos x-\sin x)^{2}+\cos ^{4} x-\sin ^{4} x=0.5 \sin 4 x$. | ## Solution.
From the condition we get
$$
\begin{aligned}
& \cos ^{2} x-2 \cos x \sin x+\sin ^{2} x+\left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)- \\
& -0.5 \sin 4 x=0 \Leftrightarrow 1-\sin 2 x+\cos 2 x-\sin 2 x \cos 2 x=0 \Leftrightarrow \\
& \Leftrightarrow(1+\cos 2 x)-(\sin 2 x+\sin 2 x... | x_{1}=\frac{\pi}{2}(2n+1);x_{2}=\frac{\pi}{4}(4k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,614 |
8.327. $\operatorname{ctg} x\left(1-\frac{1}{2} \cos 2 x\right)=1$. | ## Solution.
Domain of definition: $\sin x \neq 0$.
$$
\begin{aligned}
& \text { By the formula } \cos \alpha=\frac{1-\operatorname{tg}^{2} \frac{\alpha}{2}}{1+\operatorname{tg}^{2} \frac{\alpha}{2}}, \text { we have } \\
& \frac{1}{\operatorname{tg} x}\left(1-\frac{1-\operatorname{tg}^{2} x}{2\left(1+\operatorname{t... | \frac{\pi}{4}+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,615 |
8.328. $\cos ^{2}\left(x+40^{\circ}\right)+\cos ^{2}\left(x-40^{\circ}\right)-\sin 10^{\circ} \cos 2 x=\sin 2 x$. | ## Solution.
Reducing the degree, we have
$$
\begin{aligned}
& \frac{1}{2}\left(1+\cos \left(2 x+80^{\circ}\right)\right)+\frac{1}{2}\left(1+\cos \left(2 x-80^{\circ}\right)\right)-\sin 10^{\circ} \cos 2 x=\sin 2 x \\
& 1+\frac{1}{2}\left(\cos \left(2 x+80^{\circ}\right)+\cos \left(2 x-80^{\circ}\right)\right)-\sin 1... | \frac{\pi}{4}(4k+1),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,616 |
8.329. $2 \cos ^{2} \frac{x}{2}(1-\sin x)+\cos ^{2} x=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.329. $2 \cos ^{2} \frac{x}{2}(1-\sin x)+\cos ^{2} x=0$. | ## Solution.
The sum of two non-negative numbers is zero if and only if each of them is zero. Thus, we have:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 2 \operatorname { cos } \frac { 2 x } { 2 } ( 1 - \operatorname { sin } x ) = 0 , } \\
{ \operatorname { cos } ^ { 2 } x = 0 , }
\end{array} \Leftrightarrow \... | \frac{\pi}{2}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,617 |
8.330. $\operatorname{tg} 6 x \cos 2 x-\sin 2 x-2 \sin 4 x=0$. | Solution.
Domain of definition: $\cos 6 x \neq 0$.
Let's write the equation in the form
$\left(\frac{\sin 6 x \cos 2 x}{\cos 6 x}-\sin 2 x\right)-2 \sin 4 x=0$
$\frac{\sin 6 x \cos 2 x-\cos 6 x \sin 2 x}{\cos 6 x}-2 \sin 4 x=0 \Leftrightarrow \sin 4 x-2 \sin 4 x \cos 6 x=0$,
$\sin 4 x(1-2 \cos 6 x)=0$.
## From th... | x_{1}=\frac{\pi}{2};x_{2}=\frac{\pi}{18}(6\1),,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,618 |
8.331. $\cos 8 x+3 \cos 4 x+3 \cos 2 x=8 \cos x \cos ^{3} 3 x-0.5$. | ## Solution.
Let's rewrite the equation as
$$
\begin{aligned}
& \cos 8 x + 3 \cos 4 x + 3 \cos 2 x = 2(2 \cos x \cos 3 x)\left(2 \cos ^{2} 3 x\right) - 0.5 \Leftrightarrow \\
& \Leftrightarrow \cos 8 x + 3 \cos 4 x + 3 \cos 2 x = 2(\cos 2 x + \cos 4 x)(1 + \cos 6 x) - 0.5 \Leftrightarrow \\
& \Leftrightarrow \cos 8 x... | \frac{\pi}{30}(6k\1),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,619 |
8.332. $\operatorname{tg} x \operatorname{tg}(x+1)=1$.
8.332. $\tan x \tan(x+1)=1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0 \\ \cos (x+1) \neq 0\end{array}\right.$
## Write the equation in the form
$\frac{\sin x \sin (x+1)}{\cos x \cos (x+1)}-1=0 \Leftrightarrow \sin x \sin (x+1)-\cos x \cos (x+1)=0 \Leftrightarrow$
$\Leftrightarrow-\cos (2 x+1)=0, 2 x+1=\frac{\pi}{2}... | \frac{\pi}{4}-\frac{1}{2}+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,620 |
8.334. $2+\sin t=3 \operatorname{tg} \frac{t}{2}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
8.334. $2+\sin t=3 \operatorname{tg} \frac{t}{2}$. | ## Solution.
Domain of definition: $\cos \frac{t}{2} \neq 0$.
$$
\begin{aligned}
& \text { By the formula } \sin \alpha=\frac{2 \operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg}^{2} \frac{\alpha}{2}} \text {, we have } \\
& 2+\frac{2 \operatorname{tg} \frac{t}{2}}{1+\operatorname{tg}^{2} \frac{t}{2}}-3 \operato... | \frac{\pi}{2}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,622 |
8.335. $\operatorname{tg}\left(35^{\circ}+x\right) \operatorname{ctg}\left(10^{\circ}-x\right)=. \frac{2}{3}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos \left(35^{\circ}+x\right) \neq 0 \\ \sin \left(10^{\circ}-x\right) \neq 0\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& \operatorname{tg}\left(35^{\circ}+x\right) \operatorname{ctg}\left(90^{\circ}-\left(80^{\circ}+x\right)\right)=\frac{2... | x_{1}=-\frac{\pi}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,623 |
8.337. $\sin ^{4} 2 x+\sin ^{3} 2 x \cos 2 x-8 \sin 2 x \cos ^{3} 2 x-8 \cos ^{4} 2 x=0$. | ## Solution.
Let's write the equation as
$\sin ^{3} 2 x(\sin 2 x+\cos 2 x)-8 \cos ^{3} 2 x(\sin 2 x+\cos 2 x)=0 \Leftrightarrow$
$\Leftrightarrow(\sin 2 x+\cos 2 x)\left(\sin ^{3} 2 x-8 \cos ^{3} 2 x\right)=0$.
From this, $\sin 2 x+\cos 2 x=0$ or $\sin ^{3} 2 x-8 \cos ^{3} 2 x=0$.
1) $\operatorname{tg} 2 x=-1,2 x=... | x_{1}=-\frac{\pi}{8}+\frac{\pik}{2},x_{2}=\frac{1}{2}\operatorname{arctg}2+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,624 |
8.338. $\cos t(1-\operatorname{tg} t)(\sin t+\cos t)=\sin t$.
8.338. $\cos t(1-\tan t)(\sin t+\cos t)=\sin t$. | Solution.
Domain of definition: $\cos t \neq 0$.
Rewrite the equation as
$$
\begin{aligned}
& \cos t\left(1-\frac{\sin t}{\cos t}\right)(\sin t+\cos t)-\sin t=0 \Leftrightarrow \\
& \Leftrightarrow(\cos t-\sin t)(\cos t+\sin t)-\sin t=0 \Leftrightarrow \cos ^{2} t-\sin ^{2} t-\sin t=0 \Leftrightarrow \\
& \Leftright... | (-1)^{}\frac{\pi}{6}+\pi,\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,625 |
8.340. $1+\sin z+\cos z+\sin 2z+\cos 2z=0$. | ## Solution.
## Rewrite the equation as
$$
\begin{aligned}
& \sin ^{2} z+\cos ^{2} z+\sin z+\cos z+2 \sin z \cos z+\cos ^{2} z-\sin ^{2} z=0 \Leftrightarrow \\
& \Leftrightarrow(\sin z+\cos z)+\left(2 \sin z \cos z+2 \cos ^{2} z\right)=0 \Leftrightarrow \\
& \Leftrightarrow(\sin z+\cos z)+2 \cos z(\sin z+\cos z)=0 \L... | z_{1}=\frac{\pi}{4}(4k-1);z_{2}=\frac{2}{3}\pi(3n\1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,626 |
8.341. $\operatorname{ctg}\left(x-25^{\circ}\right)+\operatorname{tg}\left(3 x+15^{\circ}\right)=2 \sin \left(2 x-50^{\circ}\right)$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin \left(x-25^{\circ}\right) \neq 0, \\ \cos \left(3 x+15^{\circ}\right) \neq 0,\end{array} \Leftrightarrow\left\{\begin{array}{l}x-25^{\circ} \neq 180^{\circ} t, \\ 3 x+15^{\circ} \neq 90^{\circ}+180^{\circ} p,\end{array} \Leftrightarrow\right.\right.$ $\Le... | x_{1}=115+180r,r\inZ;x_{2}=-20+90n,n\inZ;x_{3}=-5+180\mathrm{},\inZ;x_{4}=55+180 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,627 |
8.342. $\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x+3 \operatorname{tg} x+3 \operatorname{ctg} x+4=0$.
8.342. $\tan^{2} x+\cot^{2} x+3 \tan x+3 \cot x+4=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0 .\end{array}\right.$
Write the equation in the form
$$
\begin{aligned}
& (\operatorname{tg} x+\operatorname{ctg} x)^{2}-2 \operatorname{tg} x \operatorname{tg} x+3(\operatorname{tg} x+\operatorname{ctg} x)+4=0 \\
& (\operatorn... | \frac{\pi}{4}(4k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,628 |
8.344. $\cos 2 x=\cos ^{2} \frac{3 x}{2}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.344. $\cos 2 x=\cos ^{2} \frac{3 x}{2}$. | ## Solution.
From the condition we have
$$
\cos 2 x=\frac{1}{2}(1+\cos 3 x)
$$
and using the formulas $\cos 2 \alpha=2 \cos ^{2} \alpha-1, \cos 3 x=4 \cos ^{3} x-3 \cos x$, we get
$$
\begin{aligned}
& 2\left(2 \cos ^{2} x-1\right)-1-\left(4 \cos ^{3} x-3 \cos x\right)=0 \Leftrightarrow \\
& \Leftrightarrow 4 \cos ^... | x_{1}=2\pik;x_{2}=\frac{\pi}{6}(6n\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,629 |
8.345. $(\tan t-\cot t+2 \tan 2 t)(1+\cos 3 t)=4 \sin 3 t$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \sin t \neq 0, \\ \cos 2 t \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
\left(\frac{\sin t}{\cos t}-\frac{\cos t}{\sin t}+2 \operatorname{tg} 2 t\right) \cdot(1+\cos 3 t)-4 \sin 3 t=0 \Leftrightarrow
$$
$$
\begin{aligned}
& \Left... | t_{1}=\frac{\pi}{5}(2k+1),k\neq5+2;t_{2}=\frac{\pi}{3}(2n+1),n\neq3+1,wherek,n,\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,630 |
8.348. $1-\sin 2x=\cos x-\sin x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
8.348. $1-\sin 2x=\cos x-\sin x$. | ## Solution.
Let's rewrite the equation as
$$
\begin{aligned}
& \cos ^{2} x-2 \sin x \cos x+\sin ^{2} x-(\cos x-\sin x)=0 \Leftrightarrow \\
& \Leftrightarrow(\cos x-\sin x)^{2}-(\cos x-\sin x)=0 \Leftrightarrow \\
& \Leftrightarrow(\cos x-\sin x)(\cos x-\sin x-1)=0
\end{aligned}
$$
From this, we have 1) $\cos x-\si... | x_{1}=\frac{\pi}{4}(4k+1);x_{2}=2\pi;x_{3}=\frac{\pi}{2}(4n-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,633 |
8.349. $\operatorname{tg}^{4} x+\operatorname{tg}^{2} x+\operatorname{ctg}^{4} x-\operatorname{ctg}^{2} x=\frac{106}{9}.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0 .\end{array}\right.$
Since $\operatorname{tg}^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{1+\cos \alpha}$ and $\operatorname{ctg}^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{1-\cos \alpha}$, we get
$$
\begin{aligned}
& \left(\frac... | x_{1}=\\frac{1}{2}\arccos\frac{-6+\sqrt{157}}{11}+\pik;x_{2}=\frac{\pi}{3}(3n\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,634 |
8.350. $\cos ^{2}\left(2 x+\frac{\pi}{3}\right)+\cos ^{2}\left(\frac{\pi}{12}-x\right)=0$. | Solution.
Let's write the equation as
$$
\begin{aligned}
& \cos ^{2}\left(2 x+\frac{\pi}{3}\right)+\cos ^{2}\left(\frac{\pi}{12}-x\right)=0 \Leftrightarrow\left\{\begin{array}{l}
\cos \left(2 x+\frac{\pi}{3}\right)=0 \\
\cos \left(x-\frac{\pi}{12}\right)=0
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\... | \frac{7\pi}{12}+\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,635 |
8.351. $3 \sqrt{3} \tan x \sin x - \cot x \cos x + 9 \sin x - 3 \sqrt{3} \cos x = 0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& \frac{3 \sqrt{3} \sin ^{2} x}{\cos x}-\frac{\cos ^{2} x}{\sin x}+9 \sin x-3 \sqrt{3} \cos x=0 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt{3} \sin ^{3} x+9 \sin ^... | x_{1}=\frac{\pi}{6}(6k+1),x_{2}=\operatorname{arctg}\frac{-2\sqrt{3}-3}{3}+\pin,x_{3}=\operatorname{arctg}\frac{-2\sqrt{3}+3}{3}+\pi,\text{where}k,n\text{}\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,636 |
8.352. $\cos 2 x-\cos x+\cos \left(x+\frac{\pi}{4}\right)+\sin \left(x+\frac{\pi}{4}\right)=\sin \frac{\pi}{4}-1$. | ## Solution.
By the formulas
$$
\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta, \sin \left(\alpha^{\circ}+\beta\right)=\sin \alpha \cos \beta+\cos \alpha \sin \beta
$$
$$
\text { and } \cos 2 \alpha=2 \cos ^{2} \alpha-1
$$
we have
$$
2 \cos ^{2} x-1-\cos x+\cos \frac{\pi}{4} \cos x-\sin \frac{\p... | x_{1}=\frac{\pi}{4}(8k\3);x_{2}=\frac{\pi}{3}(6n\1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,637 |
8.353. $\operatorname{tg}^{4} x+\operatorname{ctg}^{4} x+\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x=4$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0 .\end{array}\right.$
Since
$$
a^{4}+b^{4}=\left((a+b)^{2}-2 a b\right)^{2}-2 a^{2} b^{2} \text { and }(a+b)^{2}=(a+b)^{2}-2 a b,
$$
we have
$$
\begin{aligned}
& \left((\operatorname{tg} x+\operatorname{ctg} x)^{2}-2 \operatorna... | \frac{\pi}{4}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,638 |
8.354. $\operatorname{tg}\left(\frac{3 \pi}{2}-x\right)+\frac{\cos \left(\frac{7 \pi}{2}+x\right)}{1+\cos x}=2$. | ## Solution.
$\left\{\begin{array}{l}\sin x \neq 0 \\ \cos x \neq-1\end{array}\right.$
## We have
$$
\begin{aligned}
& \operatorname{ctg} x+\frac{\sin x}{1+\cos x}-2=0, \frac{\cos x}{\sin x}+\frac{\sin x}{1+\cos x}-2=0 \Leftrightarrow \\
& \Leftrightarrow \cos x(1+\cos x)+\sin ^{2} x-2 \sin x(1+\cos x)=0 \Leftrighta... | (-1)^{\}\frac{\pi}{6}+\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,639 |
8.355. $\operatorname{tg} x-\operatorname{tg} 2 x=\sin x$.
8.355. $\tan x-\tan 2 x=\sin x$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos 2 x \neq 0 .\end{array}\right.$
By the formula $\operatorname{tg} \alpha-\operatorname{tg} \beta=\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta}$, we have
$$
\frac{-\sin x}{\cos x \cos 2 x}-\sin x=0 \Leftrightarrow-\sin x\left(\frac... | \pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,640 |
8.357. $4 \sin ^{4} x+\cos 4 x=1+12 \cos ^{4} x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.357. $4 \sin ^{4} x+\cos 4 x=1+12 \cos ^{4} x$. | ## Solution.
Rewrite the equation as
$4\left(\sin ^{2} x\right)^{2}+\cos 2(2 x)-1-12\left(\cos ^{2} x\right)^{2}=0 \Leftrightarrow$
$\Leftrightarrow(1-\cos 2 x)^{2}+2 \cos ^{2} 2 x-1-1-3(1+\cos 2 x)^{2}=0 \Leftrightarrow \cos 2 x=-\frac{1}{2}$,
$2 x= \pm \frac{2}{3} \pi+2 \pi k, x= \pm \frac{\pi}{3}+\pi k=\frac{\pi... | \frac{\pi}{3}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,642 |
8.359. $37 \operatorname{tg} 3x = 11 \operatorname{tg} x$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 3 x \neq 0, \\ \cos x \neq 0 .\end{array}\right.$
Using the formula $\operatorname{tg} 3 \alpha=\frac{3 \operatorname{tg} \alpha-\operatorname{tg}^{3} \alpha}{1-3 \operatorname{tg}^{2} \alpha}$, rewrite the equation as
$$
37 \cdot \frac{3 \operatorname{tg}... | x_{1}=\pik,k\inZ;x_{2}=\\operatorname{arctg}5+\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,643 |
8.360. $\sqrt{2}\left(\cos ^{4} 2 x-\sin ^{4} 2 x\right)=\cos 2 x+\sin 2 x$. | ## Solution.
Let's write the equation as
$$
\begin{aligned}
& \sqrt{2}\left(\cos ^{2} 2 x+\sin ^{2} 2 x\right)\left(\cos ^{2} 2 x-\sin ^{2} 2 x\right)-(\cos 2 x+\sin 2 x)=0 \Leftrightarrow \\
& \Leftrightarrow \sqrt{2}(\cos 2 x+\sin 2 x)(\cos 2 x-\sin 2 x)-(\cos 2 x+\sin 2 x)=0 \Leftrightarrow \\
& \Leftrightarrow(\c... | x_{1}=-\frac{\pi}{8}(4k-1),k\in\mathbb{Z};x_{2}=\\frac{\pi}{6}-\frac{\pi}{8}+\pin,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,644 |
8.361. $\operatorname{tg} x-\operatorname{ctg} x=\sin ^{-1} x-\cos ^{-1} x$.
8.361. $\tan x - \cot x = \sin^{-1} x - \cos^{-1} x$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0\end{array}\right.$
## Write the equation in the form
$\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}=\frac{1}{\sin x}-\frac{1}{\cos x} \Leftrightarrow\left(\sin ^{2} x-\cos ^{2} x\right)+(\sin x-\cos x)=0$,
$(\sin x-\cos x)(\sin x+... | \frac{\pi}{4}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,645 |
8.362. $\sin ^{6} x+\cos ^{6} x=\frac{7}{16}$.
8.362. $\sin ^{6} x+\cos ^{6} x=\frac{7}{16}$. | ## Solution.
Let's write the equation as
$$
\begin{aligned}
& \left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)-\frac{7}{16}=0 \Leftrightarrow \\
& \Leftrightarrow\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-3 \sin ^{2} x \cos ^{2} x-\frac{7}{16}=0 \Leftrightarrow \\
& \Left... | \frac{\pi}{6}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,646 |
8.363. $\sin 3 x+\sin x-\sin 2 x=2 \cos x(\cos x-1)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.363. $\sin 3 x+\sin x-\sin 2 x=2 \cos x(\cos x-1)$. | ## Solution.
Since $\sin 2 \alpha=2 \sin \alpha \cos \alpha$ and $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we have
$$
\begin{aligned}
& 3 \sin x-4 \sin ^{3} x+\sin x-2 \sin x \cos x-2 \cos x(\cos x-1)=0 \Leftrightarrow \\
& \Leftrightarrow\left(4 \sin x-4 \sin ^{3} x-2 \sin x \cos x\right)-2 \cos x(\cos x-1)=... | x_{1}=\frac{\pi}{2}(2k+1),k\inZ;x_{2}=2\pin,n\inZ;x_{3}=\frac{\pi}{4}(4-1),\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,647 |
8.364. $\cos 2 x=\frac{1+\sqrt{3}}{2} \cdot(\cos x+\sin x)$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.364. $\cos 2 x=\frac{1+\sqrt{3}}{2} \cdot(\cos x+\sin x)$. | Solution.
Let's write the equation as
$\cos ^{2} x-\sin ^{2} x-\frac{1+\sqrt{3}}{2}(\cos x+\sin x)=0 \Leftrightarrow$
$\Leftrightarrow(\cos x+\sin x)(\cos x-\sin x)-\frac{1+\sqrt{3}}{2} \cdot(\cos x+\sin x)=0 \Leftrightarrow$
$\Leftrightarrow(\cos x+\sin x)\left(\cos x-\sin x-\frac{1+\sqrt{3}}{2}\right)=0$.
From t... | x_{1}=\frac{\pi}{4}(4k-1),k\inZ;x_{2}=\frac{\pi}{3}(6n-1),n\inZ;x_{3}=\frac{\pi}{6}(12-1),\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,648 |
8.365. $2(1+\sin 2 x)=\tan\left(\frac{\pi}{4}+x\right)$. | ## Solution.
Domain of definition: $\cos \left(\frac{\pi}{4}+x\right) \neq 0$.
Rewrite the equation as
$$
\begin{aligned}
& 2\left(\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x\right)-\frac{\sin \left(\frac{\pi}{4}+x\right)}{\cos \left(\frac{\pi}{4}+x\right)}=0 \Leftrightarrow \\
& \Leftrightarrow 2(\sin x+\cos x)^{2}-\fr... | x_{1}=\frac{\pi}{4}(4k-1);x_{2}=\frac{\pi}{6}(6n\1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,649 |
8.367. $\frac{\tan 2 t}{\cos ^{2} t}-\frac{\tan t}{\cos ^{2} 2 t}=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \cos 2 t \neq 0 .\end{array}\right.$
Write the equation in the form
$\frac{\sin 2 t}{\cos 2 t \cos ^{2} t}-\frac{\sin t}{\cos t \cos ^{2} 2 t}=0, \frac{\sin 2 t \cos 2 t-\sin t \cos t}{\cos ^{2} t \cos ^{2} 2 t}=0 \Leftrightarrow$ $\Leftrighta... | t_{1}=\pik,t_{2}=\\frac{\pi}{6}+\pin,k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,651 |
8.368. $\tan x + \tan 2x + \tan 3x = 0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos 2 x \neq 0, \\ \cos 3 x \neq 0 .\end{array}\right.$
Since $\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}$ and $\operatorname{tg} 3 \alpha=\frac{3 \operatorname{tg} \alpha-\operatorname{tg}^... | x_{1}=\pin,n\inZ;x_{2}=\pik\\operatorname{arctg}\frac{\sqrt{2}}{2},k\inZ;x_{3}=\frac{\pi}{3}(3\1),\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,652 |
8.369. $\operatorname{ctg} x-\operatorname{tg} x=\sin x+\cos x$.
8.369. $\cot x - \tan x = \sin x + \cos x$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos x \neq 0 .\end{array}\right.$
## Rewrite the equation as
$$
\begin{aligned}
& \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}-(\cos x+\sin x)=0 \Leftrightarrow \\
& \Leftrightarrow \cos ^{2} x-\sin ^{2} x-(\cos x+\sin x) \sin x \cos x=0 \\... | x_{1}=\frac{\pi}{4}(4k-1);x_{2}=(-1)^{n}\arcsin\frac{1-\sqrt{2}}{\sqrt{2}}+\frac{\pi}{4}(4n+1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,653 |
8.370. $\sqrt{\cos ^{2} x+\frac{1}{2}}+\sqrt{\sin ^{2} x+\frac{1}{2}}=2$. | ## Solution.
Let
$$
\left\{\begin{array} { l }
{ \sqrt { \cos ^ { 2 } x + \frac { 1 } { 2 } } = u > 0 , } \\
{ \sqrt { \sin ^ { 2 } x + \frac { 1 } { 2 } } = v > 0 . }
\end{array} \Rightarrow \left\{\begin{array}{l}
\cos ^{2} x+\frac{1}{2}=u^{2}>0, \\
\sin ^{2} x+\frac{1}{2}=v^{2}>0
\end{array} \Rightarrow u^{2}+v^{... | \frac{\pi}{4}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,654 |
8.371. $\sin 3 x=a \sin x$.
8.371. $\sin 3 x=a \sin x$. | ## Solution.
From the formula $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we have
$$
\begin{aligned}
& 3 \sin x-4 \sin ^{3} x-a \sin x=0,4 \sin ^{3} x+(a-3) \sin x=0 \\
& \sin x\left(4 \sin ^{2} x+a-3\right)=0
\end{aligned}
$$
From this,
1) $\sin x=0, x_{1}=\pi k, k \in Z$;
2) $4 \sin ^{2} x+a-3=0, \sin ^{2} ... | x_{1}=\pik,x_{2}=\\frac{1}{2}\arccos\frac{-1}{2}+\pin,k,n\inZ,\in[-1;3] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,655 |
8.372. $\cos 3 x=m \cos x$.
8.372. $\cos 3 x=m \cos x$. | ## Solution.
From the formula $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$, we have $4 \cos ^{3} x-3 \cos x-m \cos x=0, \cos x\left(4 \cos ^{2} x-3-m\right)=0$.
## Hence
1) $\cos x=0, x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$;
2) $4 \cos ^{2} x-3-m=0 \Leftrightarrow 2(1+\cos 2 x)=m+3, \cos 2 x=\fr... | x_{1}=\frac{\pi}{2}(2k+1),\,x_{2}=\\frac{1}{2}\arccos\frac{+1}{2}+\pin,\,\in[-3;1],\,k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,656 |
8.373. $\operatorname{tg} x+\operatorname{tg} \alpha+1=\operatorname{tg} x \operatorname{tg} \alpha$.
8.373. $\tan x+\tan \alpha+1=\tan x \tan \alpha$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos \alpha \neq 0 .\end{array}\right.$
Write the equation in the form
$\operatorname{tg} x+\operatorname{tg} \alpha=-(1-\operatorname{tg} x \operatorname{tg} \alpha) \Rightarrow \frac{\operatorname{tg} x+\operatorname{tg} \alpha}{1-\operatorn... | -\alpha+\frac{\pi}{4}(4k-1)for\alpha\neq\frac{\pi}{4},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,657 |
8.374. $12 \sin x + 4 \sqrt{3} \cos (\pi + x) = m \sqrt{3}$. | ## Solution.
From the condition we have
$$
\begin{aligned}
& 12 \sin x-4 \sqrt{3} \cos x=8 \sqrt{3}\left(\frac{\sqrt{3}}{2} \sin x-\frac{1}{2} \cos x\right)=8 \sqrt{3}\left(\sin 60^{\circ} \sin x-\right. \\
& \left.-\cos 60^{\circ} \cos x\right)=-8 \sqrt{3} \cos \left(60^{\circ}+x\right)=8 \sqrt{3} \sin \left(x-30^{\... | (-1)^{n}\arcsin\frac{}{8}+\frac{\pi}{6}(6n+1),where-8\leq\leq8,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,658 |
8.377. $2^{\sin ^{2} x}+4 \cdot 2^{\cos ^{2} x}=6$.
8.377. $2^{\sin ^{2} x}+4 \cdot 2^{\cos ^{2} x}=6$. | ## Solution.
Let's rewrite the equation as
$$
2^{\sin ^{2} x}+4 \cdot 2^{1-\sin ^{2} x}-6=0 \Leftrightarrow\left(2^{\sin ^{2} x}\right)^{2}-6\left(2^{\sin ^{2} x}\right)+8=0
$$
Solving it as a quadratic equation in terms of $2^{\sin ^{2} x}$, we get
1) $2^{\sin ^{2} x}=2, \sin ^{2} x=1, \sin x= \pm 1, x=\frac{\pi}{... | \frac{\pi}{2}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,659 |
8.382. $1+2^{\operatorname{tg} x}=3 \cdot 4^{-\frac{1}{\sqrt{2}} \sin \left(\frac{\pi}{4}-x\right) \cos ^{-1} x}$. | Solution.
Domain of definition: $\cos x \neq 0$.
From the condition we have
$$
\begin{aligned}
& 2\left(\frac{\sin \frac{\pi}{4} \cos x-\cos \frac{\pi}{4} \sin x}{\sqrt{2} \cos x}\right) \\
& 1+2^{\operatorname{tg} x}=3 \cdot 2^{\operatorname{tg} x}=2^{\frac{\cos x-\sin x}{\cos x}} \Leftrightarrow \\
& \Leftrightarr... | \frac{\pi}{4}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,662 |
8.383. $\log _{\cos x} 4 \cdot \log _{\cos ^{2} x} 2=1$. | ## Solution.
Domain of definition: $0<\cos x<1$.
Switch to the base $\cos x$. We get
$\frac{1}{2} \log _{\cos x} 2^{2} \cdot \log _{\cos x} 2=1, \log _{\cos x}^{2} 2=1$.
From here
1) $\log _{\cos x} 2=1 \Leftrightarrow \cos x=2, \varnothing$
2) $\log _{\cos x} 2=-1, \frac{1}{\cos x}=2, \cos x=\frac{1}{2}, x= \pm \... | \frac{\pi}{3}(6k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,663 |
8.384. $\log _{\sin x} 4 \cdot \log _{\sin ^{2} x} 2=4$. | Solution.
Domain of definition: $0<\sin x<1$.
Switch to the base $\sin x$. We have
$$
\log _{\sin x} 2^{2} \cdot \frac{1}{2} \log _{\sin x} 2=4, \log _{\sin x}^{2} 2=4
$$
## Hence
1) $\log _{\sin x} 2=2, \sin ^{2} x=2, \varnothing$;
2) $\log _{\sin x} 2=-2, \frac{1}{\sin ^{2} x}=2, \sin ^{2} x=\frac{1}{2}, \sin x=... | (-1)^{k}\frac{\pi}{4}+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,664 |
8.385. $3\left(\log _{2} \sin x\right)^{2}+\log _{2}(1-\cos 2 x)=2$. | ## Solution.
Domain of definition: $\sin x>0$.
From the condition we have
$$
\begin{aligned}
& 3\left(\log _{2} \sin x\right)^{2}+\log _{2}\left(1-1+2 \sin ^{2} x\right)-2=0 \Leftrightarrow \\
& \Leftrightarrow 3\left(\log _{2} \sin x\right)^{2}+\log _{2}\left(2 \sin ^{2} x\right)-2=0 \Leftrightarrow \\
& \Leftright... | (-1)^{k}\frac{\pi}{6}+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,665 |
8.386. Given $(1+\operatorname{tg} x)(1+\operatorname{tg} y)=2$. Find $x+y$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos y \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& 1+\operatorname{tg} x+\operatorname{tg} y+\operatorname{tg} x \operatorname{tg} y=2, \operatorname{tg} x+\operatorname{tg} y=1-\operatorname{tg} x \operatorname... | x+\frac{\pi}{4}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,666 |
8.388. One of the angles of a right triangle satisfies the equation $\sin ^{3} x+\sin x \sin 2 x-3 \cos ^{3} x=0$. Show that the triangle is isosceles. | ## Solution.
Let's write the equation as
$$
\sin ^{3} x+\sin x \cdot 2 \sin x \cos x-3 \cos ^{3} x=0 \Leftrightarrow
$$
$\Leftrightarrow \sin ^{3} x+2 \sin ^{2} x \cos x-3 \cos ^{3} x=0 \Leftrightarrow \operatorname{tg}^{3} x+2 \operatorname{tg}^{2} x-3=0 \Leftrightarrow$
$\Leftrightarrow \operatorname{tg}^{3} x-\o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,668 |
8.389. Show that there does not exist a triangle, each angle of which satisfies the equation
\[
(3 \cos x-2)\left(14 \sin ^{2} x+\sin 2 x-12\right)=0
\] | ## Solution.
The equation is equivalent to the system of equations
1) $3 \cos x-2=0$ or 2) $14 \sin ^{2} x+\sin 2 x-12=0$.
2) $\cos x=\frac{2}{3}$. Then $\sin x=+\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}, \operatorname{tg} x=\frac{\sin x}{\cos x}=\frac{\sqrt{5}}{2}$.
3) $14 \sin ^{2} x+2 \sin x \cos x-12\left(\sin ^{2}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,669 |
8.390. Show that there exist triangles in which each angle satisfies the equation $(65 \sin x-56)\left(80-64 \sin x-65 \cos ^{2} x\right)=0$. Find these angles. | ## Solution.
The equation is equivalent to a system of two equations:
1) $65 \sin x - 56 = 0$ or 2) $80 - 64 \sin x - 65 \cos^2 x = 0$.
2) $\sin x = \frac{56}{65}$
3) $80 - 64 \sin x - 65(1 - \sin^2 x) = 0 \Leftrightarrow 65 \sin^2 x - 64 \sin x + 15 = 0$, from which $\sin x = \frac{5}{13}$ or $\sin x = \frac{3}{5}$.... | x_1=\arcsin\frac{3}{5},x_2=\arcsin\frac{5}{13},x_3=\pi-\arcsin\frac{56}{65} | Geometry | proof | Yes | Yes | olympiads | false | 48,670 |
8.391. Show that a triangle, each of whose angles satisfies the equation $3 \operatorname{tg} x-3 \operatorname{tg}\left(\frac{x}{2}\right)-2 \sqrt{3}=0$, is equilateral. | ## Solution.
From the condition we have
$$
\begin{aligned}
& \frac{6 \operatorname{tg} \frac{x}{2}}{1-\operatorname{tg}^{2} \frac{x}{2}}-3 \operatorname{tg} \frac{x}{2}-2 \sqrt{3}=0 \Leftrightarrow 3 \operatorname{tg}^{3} \frac{x}{2}+2 \sqrt{3} \operatorname{tg}^{2} \frac{x}{2}+3 \operatorname{tg} \frac{x}{2}-2 \sqrt... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,671 |
8.392. Find $\sin \alpha$, if $\cos \alpha=\operatorname{tg} \beta, \cos \beta=\operatorname{tg} \gamma, \cos \gamma=\operatorname{tg} \alpha\left(0<\alpha<\frac{\pi}{2}\right.$, $\left.0<\beta<\frac{\pi}{2}, 0<\gamma<\frac{\pi}{2}\right)$. | ## Solution.
From the condition we have
$$
\left\{\begin{array} { l }
{ \operatorname { cos } \alpha = \operatorname { tan } \beta , } \\
{ \operatorname { cos } \beta = \operatorname { tan } \gamma , } \\
{ \operatorname { cos } \gamma = \operatorname { tan } \alpha , }
\end{array} \Leftrightarrow \left\{\begin{arr... | \sin\alpha=\frac{\sqrt{5}-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,672 |
8.393. Find the angles $\alpha, \beta$ and $\gamma$ of the first quadrant, if it is known that they form an arithmetic progression with a common difference of $\frac{\pi}{12}$, and their tangents form a geometric progression | Solution.
Let $\alpha, \beta$ and $\gamma$ be members of an arithmetic progression, $d=\frac{\pi}{12}$; $\operatorname{tg} \alpha, \operatorname{tg} \beta, \operatorname{tg} \gamma$ be members of a geometric progression; $\beta=\alpha+\frac{\pi}{12}, \gamma=\alpha+\frac{\pi}{6}$.
By the property of members of a geome... | \frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,673 |
8.394. $\left\{\begin{array}{l}\sin x+\cos y=0, \\ \sin ^{2} x+\cos ^{2} y=\frac{1}{2} .\end{array}\right.$ | ## Solution.
Let's write the given system of equations in the form
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ \sin x + \cos y = 0 , } \\
{ (\sin x + \cos y)^2 - 2 \sin x \cos y = \frac { 1 } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\sin x + \cos y = 0, \\
\sin x \cos y = -\frac{1}{4}
\end{a... | x_{1}=(-1)^{k+1}\frac{\pi}{6}+\pik,y_{1}=\\frac{\pi}{3}+2\pin,\quadx_{2}=(-1)^{k}\frac{\pi}{6}+\pik,\quady_{2}=\\frac{2}{3}\pi+2\pin,\text{where}k\text{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,674 |
8.395. $\left\{\begin{array}{l}9^{2 \operatorname{tg} x+\cos y}=3 \\ 9^{\cos y}-81^{\operatorname{tg} x}=2 .\end{array}\right.$ | ## Solution.
Domain of definition: $\cos x \neq 0$.
From the condition we have
$\left\{\begin{array}{l}81^{\operatorname{tg} x} \cdot 9^{\cos y}=3, \\ 9^{\cos y}-81^{\operatorname{tg} x}=2,\end{array} \Rightarrow 9^{\cos y}=2+81^{\operatorname{tg} x}, 81^{\operatorname{tg} x} \cdot\left(2+81^{\operatorname{tg} x}\ri... | \pik_{1},\\frac{\pi}{3}+2\pik_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,675 |
8.396. $\left\{\begin{array}{l}x-y=\frac{5 \pi}{3} \\ \sin x=2 \sin y .\end{array}\right.$ | ## Solution.
From the condition we have
$$
\begin{aligned}
& x=\frac{5 \pi}{3}+y, \sin \left(\frac{5 \pi}{3}+y\right)-2 \sin y=0,-\sin \left(\frac{\pi}{3}-y\right)-2 \sin y=0 \Leftrightarrow \\
& \Leftrightarrow-\sin \frac{\pi}{3} \cos y+\cos \frac{\pi}{3} \sin y-2 \sin y=0 \Leftrightarrow \sqrt{3} \cos y+3 \sin y=0 ... | \frac{\pi}{2}(2k+3),\frac{\pi}{6}(6k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,676 |
8.397. $\left\{\begin{array}{l}\sin x \cos y=0.25, \\ \sin y \cos x=0.75\end{array}\right.$ | ## Solution.
Let's add and subtract the first and second equations of the system. We get $\left\{\begin{array}{l}\sin x \cos y+\sin y \cos x=1, \\ \sin x \cos y-\sin y \cos x=-\frac{1}{2},\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin (x+y)=1, \\ \sin (x-y)=-\frac{1}{2},\end{array} \Leftrightarrow\right.\right... | x_{1}=\frac{\pi}{6}+\pi(k_{1}-k_{2}),y_{1}=\frac{\pi}{3}+\pi(k_{1}+k_{2});x_{2}=-\frac{\pi}{6}+\pi(k_{1}-k_{2}),y_{2}=\frac{2}{3}\pi+\pi(k_{1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,677 |
8.398. $\left\{\begin{array}{l}x-y=-\frac{1}{3}, \\ \cos ^{2} \pi x-\sin ^{2} \pi y=\frac{1}{2} .\end{array}\right.$ | ## Solution.
Let's rewrite the second equation of the system as
$$
\begin{aligned}
& \frac{1}{2}(1+\cos 2 \pi x)-\frac{1}{2}(1-\cos 2 \pi y)=\frac{1}{2}, \cos 2 \pi x+\cos 2 \pi y=1 \Leftrightarrow \\
& \Leftrightarrow 2 \cos \frac{2 \pi x+2 \pi y}{2} \cos \frac{2 \pi x-2 \pi y}{2}=1, 2 \cos \pi(x+y) \cos \pi(x-y)=1 ... | k-\frac{1}{6},k+\frac{1}{6},k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,678 |
8.399. $\left\{\begin{array}{l}x+y=\frac{\pi}{4} \\ \tan x \tan y=\frac{1}{6}\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos y \neq 0 .\end{array}\right.$
From the condition we have
$$
\begin{aligned}
& x=\frac{\pi}{4}-y, \operatorname{tg}\left(\frac{\pi}{4}-y\right) \operatorname{tg} y=\frac{1}{6} \Rightarrow \frac{\left(\operatorname{tg} \frac{\pi}{4}-\operat... | {\begin{array}{}{x_{1}=\operatorname{\frac{1}{3}+\pik,}\{y_{1}=\frac{\pi}{4}-\operatorname{\frac{1}{3}-\pik;}\end{pmatrix}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,679 |
8.400. $\left\{\begin{array}{l}\sqrt{2} \sin x=\sin y, \\ \sqrt{2} \cos x=\sqrt{3} \cos y .\end{array}\right.$ | ## Solution.
By squaring both equations of the system, we get
$$
\begin{aligned}
& \left\{\begin{array}{l}
2 \sin ^{2} x=\sin ^{2} y, \\
2 \cos ^{2} x=3 \cos ^{2} y .
\end{array} \Rightarrow 2=\sin ^{2} y+3 \cos ^{2} y, 2 \sin ^{2} y+3\left(1-\sin ^{2} y\right)\right. \\
& \sin ^{2} y=\frac{1}{2}, \sin y= \pm \frac{\... | \\frac{\pi}{6}+\pik_{2},\\frac{\pi}{4}+\pik_{1};k_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,680 |
8.401. $\left\{\begin{array}{l}\tan \frac{x}{2}+\tan \frac{y}{2}=2 \\ \cot x+\cot y=-1.8 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \sin y \neq 0 .\end{array}\right.$
Using the formula $\operatorname{ctg} \alpha=\frac{1-\operatorname{tg}^{2} \frac{\alpha}{2}}{2 \operatorname{tg} \frac{\alpha}{2}}$, we can write the system of equations as
$$
\begin{aligned}
& \left\{\beg... | x_{1}=-2\operatorname{arctg}\frac{1}{2}+2\pik_{1},y_{1}=2\operatorname{arctg}\frac{5}{2}+2\pik_{2};x_{2}=2\operatorname{arctg}\frac{5}{2}+2\pik_{1},y_{2}=-2\operatorname{arctg} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,681 |
8.402. $\left\{\begin{array}{l}2^{\cos x}+2^{\cos ^{-1} y}=5, \\ 2^{\cos x} \cdot 2^{\cos ^{-1} y}=4 .\end{array}\right.$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.402. $\left\{\begin{array}{l}2^{\cos x}+2^{\cos ^{-1}... | ## Solution.
From the condition we have
$$
2^{\cos ^{-1} y}=5-2^{\cos x}, 2^{\cos x} \cdot\left(5-2^{\cos x}\right)=4,\left(2^{\cos x}\right)^{2}-5\left(2^{\cos x}\right)+4=0
$$
Solving this equation as a quadratic in terms of $2^{\cos x}$, we get
$$
2^{\cos x}=1, \cos x=0, x=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1... | \frac{\pi}{2}(2k+1),\frac{\pi}{3}(6k_{1}\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,682 |
8.403. $\left\{\begin{array}{l}\sin x \sin y=0.75, \\ \tan x \tan y=3 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos y \neq 0 .\end{array}\right.$
Rewrite the second equation of the system as
$$
\frac{\sin x \sin y}{\cos x \cos y}=3 \Leftrightarrow \frac{0.75}{\cos x \cos y}=3, \cos x \cos y=0.25
$$
Then the given system has the form
$\left\{\begin{ar... | \\frac{\pi}{3}+\pi(k_{1}+k_{2}),\\frac{\pi}{3}+\pi(k_{2}-k_{1}),k_{1}\text{}k_{2}\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,683 |
8.404. $\left\{\begin{array}{l}\cos ^{2} x+\cos ^{2} y=0.25, \\ x+y=\frac{5 \pi}{6} .\end{array}\right.$ | ## Solution.
Let's write the first equation as
$$
\frac{1}{2}(1+\cos 2 x)+\frac{1}{2}(1+\cos 2 y)=0.25, \cos 2 x+\cos 2 y=-\frac{3}{2} \Leftrightarrow
$$
$\Leftrightarrow 2 \cos (x+y) \cos (x-y)=-\frac{3}{2} \Leftrightarrow 2 \cos \frac{5 \pi}{6} \cos (x-y)=-\frac{3}{2}$, $\cos (x-y)=\frac{\sqrt{3}}{2}, x-y= \pm \fr... | x_{1}=\frac{\pi}{2}(2k+1),y_{1}=\frac{\pi}{3}(1-3k);x_{2}=\frac{\pi}{3}(3k+1),y_{2}=\frac{\pi}{2}(1-2k),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,684 |
9.097. Prove that if $a$ is any real number, then the inequality $\frac{a^{2}+a+2}{\sqrt{a^{2}+a+1}} \geq 2$ holds. | Solution.
We have
$$
\frac{a^{2}+a+2}{\sqrt{a^{2}+a+1}}-2 \geq 0 \Leftrightarrow \frac{\left(a^{2}+a+1\right)-2 \sqrt{a^{2}+a+1}+1}{\sqrt{a^{2}+a+1}} \geq 0
$$
$$
\frac{\left(\sqrt{a^{2}+a+1}-1\right)^{2}}{\sqrt{a^{2}}+a+1} \geq 0, \text{ the inequality is true, since }\left(\sqrt{a^{2}+a+1}-1\right)^{2} \geq 0
$$
... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 48,687 |
9.101. For what values of $p$ are both roots of the quadratic trinomial $x^{2}+2(p+1) x+9 p-5$ negative? | ## Solution.
According to Vieta's theorem, the roots of the quadratic trinomial $A x^{2}+B x+C$ exist and are both negative if and only if
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ D = B ^ { 2 } - 4 A C \geq 0 , } \\
{ \frac { B } { A } > 0 , } \\
{ \frac { C } { A } > 0 }
\end{array} \Leftrightarrow \left\{... | p\in(\frac{5}{9};1]\cup[6;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,688 |
9.102. For what values of $n$ are both roots of the equation $(n-2) x^{2}-$ $-2 n x+n+3=0$ positive? | ## Solution.
According to Vieta's theorem, the roots of the quadratic trinomial $A x^{2}+B x+C$ exist and are both positive if and only if
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ D = B ^ { 2 } - 4 A C \geq 0 , } \\
{ \frac { B } { A } < 0, } \\
{ \frac { C } { A } > 0 }
\end{array} \Leftrightarrow \left\{\... | n\in(-\infty;-3)\cup(2;6] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,689 |
9.103. For what values of $m$ are the roots of the equation $4 x^{2}-(3 m+1) x-m-2=0$ contained in the interval between -1 and 2? | ## Solution.
The roots of the equation $a x^{2}+b x+x=0$ are contained in the interval $(\lambda, \delta)$, i.e. $\lambda 0 , } \\
{ a ( a \lambda ^ { 2 } + b \lambda + c ) > 0 , } \\
{ \lambda 0, \\
4(-1)^{2}-(3 m+1)(-1)-m-2>0, \\
-1 0 , } \\
{ 2 m + 3 > 0 , } \\
{ 3 m - 1 5 0 }
\end{array} \Leftrightarrow \left\{\b... | \in(-\frac{3}{2};\frac{12}{7}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,690 |
9.104. For what values of $a$ does the quadratic trinomial $a x^{2}-7 x+4 a$ take negative values for any real values of $x$? | Solution.
The quadratic trinomial $a x^{2}-7 x+4 a\frac{49}{16}, \\ a<0\end{array} \Leftrightarrow\left(a<-\frac{7}{4}\right)\right.\right.$.
Answer: $a \in\left(-\infty ;-\frac{7}{4}\right)$. | \in(-\infty;-\frac{7}{4}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,691 |
9.105. Find the integers $x$, satisfying the inequality $\left|\frac{2}{x-13}\right|>\frac{8}{9}$. | Solution.
From the condition, we have $|x-13|<\frac{9}{4}, x \neq 13$, from which $-\frac{9}{4}<x-3<\frac{9}{4}, \frac{43}{4}<x<\frac{61}{4}$, $x \neq 13 ; x \in\left(\frac{43}{4} ; 13\right) \cup\left(13 ; \frac{61}{4}\right)$. The integer solutions from the union of these intervals will be $x_{1}=11 ; x_{2}=12 ; x_{... | x_{1}=11;x_{2}=12;x_{3}=14;x_{4}=15 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,692 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.