problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
3.463 Find $\operatorname{ctg} \frac{x}{2}$, if it is known that $\sin x-\cos x=\frac{1+2 \sqrt{2}}{3}$. | Solution. $\sin x-\cos x=\frac{2 \tan \frac{x}{2}}{1+\tan^{2} \frac{x}{2}}-\frac{1-\tan^{2} \frac{x}{2}}{1+\tan^{2} \frac{x}{2}}=\frac{\tan^{2} \frac{x}{2}+2 \tan \frac{x}{2}-1}{1+\tan^{2} \frac{x}{2}}$.
Let $y=\tan \frac{x}{2}$, then $\frac{y^{2}+2 y-1}{1+y^{2}}=\frac{1+2 \sqrt{2}}{3} \Leftrightarrow(\sqrt{2}-1) \cdo... | \cot\frac{x}{2}=\frac{\sqrt{2}}{2},\cot\frac{x}{2}=3-2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,894 |
3.464 Prove that if $\sin \alpha=\frac{1}{3}, \sin \beta=\frac{1}{3 \sqrt{11}}, \sin \gamma=\frac{3}{\sqrt{11}}$, angles $\alpha, \beta, \gamma$ are acute and positive, then $\alpha+\beta+\gamma=\frac{\pi}{2}$. | Solution. Obviously, $\cos \alpha=\frac{2 \sqrt{2}}{3}, \cos \beta=\frac{\sqrt{98}}{3 \sqrt{11}}$. Since $\frac{1}{3}<\frac{1}{\sqrt{2}}, \frac{1}{3 \sqrt{11}}<\frac{1}{\sqrt{2}}$, then $\alpha<\frac{\pi}{4}, \quad \beta<\frac{\pi}{4}, \quad 0<\alpha+\beta<\frac{\pi}{2}, \quad 0<\frac{\pi}{2}-\gamma<\frac{\pi}{2}$. Fro... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,895 |
3.465 Given that $\operatorname{tg} \alpha=m$, find
$A=\sin ^{2}\left(\frac{\pi}{4}+\alpha\right)-\sin ^{2}\left(\frac{\pi}{6}-\alpha\right)-\cos \frac{5 \pi}{12} \cdot \sin \left(\frac{5 \pi}{12}-2 \alpha\right)$ | Solution.
$$
\begin{aligned}
& A=\frac{1-\cos \left(\frac{\pi}{2}+2 \alpha\right)}{2}-\frac{1-\cos \left(\frac{\pi}{3}-2 \alpha\right)}{2}-\frac{-\sin 2 \alpha+\sin \left(\frac{5 \pi}{6}-2 \alpha\right)}{2}= \\
& =\frac{1}{2}\left[2 \sin 2 \alpha+\cos \left(\frac{\pi}{3}-2 \alpha\right)-\sin \left(\frac{5 \pi}{6}-2 \a... | \frac{2}{1+^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,896 |
3.467 Given: $\cos 2 \alpha=m$. Find $A=\cos ^{8} \alpha-\sin ^{8} \alpha$. | ## Solution.
$A=\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right) \cdot\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right) \cdot\left(\cos ^{4} \alpha+\sin ^{4} \alpha\right)=\cos 2 \alpha \cdot\left(\cos ^{4} \alpha+\right.$ $\left.+2 \cos ^{2} \alpha \cdot \sin ^{2} \alpha+\sin ^{4} \alpha-2 \cos ^{2} \alpha \cdot \sin ^{2}... | \frac{(1+^{2})}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,897 |
3.468 Given: $\sin \alpha-\cos \alpha=$. Find $A=\frac{\sin 4 \alpha+\sin 10 \alpha-\sin 6 \alpha}{\cos 2 \alpha+1-2 \sin ^{2} 4 \alpha}$. | Solution. $A=\frac{(\sin 4 \alpha+\sin 10 \alpha)-\sin 6 \alpha}{\cos 2 \alpha+\cos 8 \alpha}=\frac{2 \sin 7 \alpha \cdot \cos 3 \alpha-2 \sin 3 \alpha \cdot \cos 3 \alpha}{2 \cos 3 \alpha \cdot \cos 5 \alpha}=$
$=\frac{\sin 7 \alpha-\sin 3 \alpha}{\cos 5 \alpha}=2 \sin 2 \alpha . \quad(\sin \alpha-\cos \alpha)^{2}=m^... | 2(1-^{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,898 |
3.469 Given: $\cos \left(x-\frac{3 \pi}{2}\right)=-\frac{4}{5}, 0<x<\frac{\pi}{2}$. Find $A=\sin \frac{x}{2} \cdot \cos \frac{5 x}{2}$. | Solution. $A=\frac{-\sin 2 x+\sin 3 x}{2}=\frac{3 \sin x-4 \sin ^{3} x-2 \sin x \cdot \cos x}{2}$.
Since $\sin x=\frac{4}{5}, \cos x=\frac{3}{5}$, then $A=-\frac{38}{125}$.
Answer: $A=-\frac{38}{125}$. | -\frac{38}{125} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,899 |
3.470 Given: $A, B, C$ - angles of a certain triangle. Prove: $\sin ^{2} A+\sin ^{2} B+\sin ^{2} C-2 \cos A \cdot \cos B \cdot \cos C=2$. | Solution. $\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=\frac{1-\cos 2 A}{2}+\frac{1-\cos 2 B}{2}+1-\cos ^{2} C=$ $=2-\frac{1}{2}(\cos 2 A+\cos 2 B)-\cos ^{2}(A+B)=2-\cos (A+B) \cdot \cos (A-B)-\cos ^{2}(A+B)=$ $=2-\cos (A+B) \cdot(\cos (A-B)+\cos (A+B))=2+2 \cos C \cdot \cos B \cdot \cos A$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,900 |
3.471 Prove that if $\cos 2 \alpha=\cos 2 \beta \cdot \cos 2 \gamma$, then $1+\operatorname{ctg}(\alpha+\beta) \cdot \operatorname{ctg}(\alpha-\beta)=\frac{1}{\sin ^{2} \gamma}$. | Solution. $1+\operatorname{ctg}(\alpha+\beta) \cdot \operatorname{ctg}(\alpha-\beta)=\frac{\sin (\alpha+\beta) \cdot \sin (\alpha-\beta)+\cos (\alpha+\beta) \cdot \cos (\alpha-\beta)}{\sin (\alpha+\beta) \cdot \sin (\alpha-\beta)}=$ $=\frac{2 \cos (\alpha+\beta-(\alpha-\beta))}{\cos 2 \beta-\cos 2 \alpha}=\frac{2 \cos ... | \frac{1}{\sin^{2}\gamma} | Algebra | proof | Yes | Yes | olympiads | false | 51,901 |
3.472 Given: $\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}$ - angles of a certain triangle. Prove:
$\sin (2 n+1) A+\sin (2 n+1) B+\sin (2 n+1) C=(-1)^{n} \cdot 4 \cos \frac{2 n+1}{2} A \times$
$\times \cos \frac{2 n+1}{2} B \cdot \cos \frac{2 n+1}{2} C$, where $n-$ is an integer. | Solution. $[\sin (2 n+1) A+\sin (2 n+1) B]+\sin (2 n+1)(\pi-(A+B))=$ $=2 \sin \frac{2 n+1}{2}(A+B) \times \cos \frac{2 n+1}{2}(A-B)+2 \sin \frac{2 n+1}{2}(A+B) \cdot \cos \frac{2 n+1}{2}(A+B)=$ $2 \sin \frac{2 n+1}{2}(A+B) \cdot\left(\cos \frac{2 n+1}{2}(A-B)+\cos \frac{2 n+1}{2}(A+B)\right)=4 \sin \frac{2 n+1}{2}(\pi-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,902 |
3.473 Given: $\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}$ - interior angles of a triangle. Prove that $\sin 2 n A+\sin 2 n B+\sin 2 n C=(-1)^{n+1} \cdot 4 \sin n A \cdot \sin n B \cdot \sin n C$, where $n$ is an integer. | Solution. $(\sin 2 n A+\sin 2 n B)+\sin 2 n(\pi-(A+B))=2 \sin n(A+B) \cdot \cos n(A-B)-$
$-2 \sin n(A+B) \cdot \cos n(A+B)=2 \sin n(\pi-C) \cdot[\cos n(A-B)-\cos n(A+B)]=$ $=-4 \sin (\pi n+n C) \cdot \sin n A \cdot \sin n B=(-1)^{n+1} \cdot 4 \sin n C \cdot \sin n B \cdot \sin n A$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,903 |
### 3.475 Prove that the expression
$A=4 \cos \alpha \cdot \cos \varphi \cdot \cos (\alpha-\varphi)+2 \sin ^{2}(\alpha-\varphi)-\cos 2 \varphi$
does not depend on $\varphi$. | ## Solution.
$A=2 \cos (\alpha-\varphi) \cdot(\cos (\alpha-\varphi)+\cos (\alpha+\varphi))+2-2 \cos ^{2}(\alpha-\varphi)-\cos 2 \varphi=$ $=2 \cos (\alpha-\varphi) \cdot \cos (\alpha+\varphi)+2-\cos 2 \varphi=2+\cos 2 \alpha$. | 2+\cos2\alpha | Algebra | proof | Yes | Yes | olympiads | false | 51,905 |
### 3.476 Find the maximum value of the expression
$A=\sin ^{2}\left(\frac{15 \pi}{8}-4 \alpha\right)-\sin ^{2}\left(\frac{17 \pi}{8}-4 \alpha\right)$ for $0 \leq \alpha \leq \frac{\pi}{8}$. | Solution.
$A=\frac{1-\cos \left(\frac{15 \pi}{4}-8 \alpha\right)}{2}-\frac{1-\cos \left(\frac{17 \pi}{4}-8 \alpha\right)}{2}=\frac{1}{2}\left(\cos \left(\frac{\pi}{4}-8 \alpha\right)-\cos \left(\frac{\pi}{4}+8 \alpha\right)\right)=$
$=\sin \frac{\pi}{4} \cdot \sin 8 \alpha=\frac{\sin 8 \alpha}{\sqrt{2}}$. Since $0 \l... | \frac{1}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,906 |
3.477 Find the minimum value of the expression
$$
A=\frac{\operatorname{ctg} 2 \alpha-\operatorname{tg} 2 \alpha}{1+\sin \left(\frac{5 \pi}{2}-8 \alpha\right)} \text { for } 0<\alpha<\frac{\pi}{8}
$$ | ## Solution.
$\operatorname{ctg} 2 \alpha-\operatorname{tg} 2 \alpha=\frac{\cos ^{2} 2 \alpha-\sin ^{2} 2 \alpha}{\sin 2 \alpha \cdot \cos 2 \alpha}=\frac{2 \cos 4 \alpha}{\sin 4 \alpha} \Rightarrow A=\frac{2 \cos 4 \alpha}{\sin 4 \alpha \cdot(1+\cos 8 \alpha)}=$ $=\frac{2 \cos 4 \alpha}{2 \sin 4 \alpha \cdot \cos ^{2... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,907 |
3.478 Prove: for one of the angles in triangle $ABC$ to be $60^{\circ}$, it is necessary and sufficient that $\sin 3A + \sin 3B + \sin 3C = 0$. | ## Solution.
1) Let $C=60^{\circ} \Rightarrow A+B=120^{\circ} \Rightarrow \sin 3 A+\sin 3 B+\sin 180^{\circ}=$ $=2 \sin \frac{3}{2}(A+B) \cdot \cos \frac{3}{2}(A-B)=0$.
2) Let $\sin 3 A+\sin 3 B+\sin 3 C=0 \Rightarrow 2 \sin \frac{3}{2}(A+B) \cdot \cos \frac{3}{2}(A-B)+$ $+\sin 3\left(180^{\circ}-(A+B)\right)=0 \Right... | proof | Geometry | proof | Yes | Yes | olympiads | false | 51,908 |
3.479 Prove: for one of the angles $A$, $B$, $C$ in a triangle to be equal to $36^{\circ}$ or $108^{\circ}$, it is sufficient that $\sin 5 A+\sin 5 B+$ $\mathbf{\sin 5 C}=\mathbf{0}$. | Solution. $(\sin 5 A+\sin 5 B)+\sin 5\left(180^{\circ}-(A+B)\right)=0 \Rightarrow 2 \sin \frac{5}{2}(A+B) \cdot \cos \frac{5}{2}(A-B)+$ $+2 \sin \frac{5}{2}(A+B) \cdot \cos \frac{5}{2}(A+B)=0 \Rightarrow \sin \frac{5}{2}(A+B) \cdot\left(\cos \frac{5}{2}(A-B)+\right.$
$\left.+\cos \frac{5}{2}(A+B)\right)=0 \Rightarrow \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,909 |
3.480 Prove: for one of the angles of a triangle to be equal to $36^{\circ}$ or $108^{\circ}$, it is necessary that $\sin 5 A+\sin 5 B+\sin 5 C=0$. | Solution. Let $C=36^{\circ} \Rightarrow 5 C=180^{\circ}, A+B=144^{\circ} \Rightarrow$
$\Rightarrow L=\sin 5 A+\sin 5 B+\sin 5 C=2 \sin \frac{5}{2}(A+B) \cdot \cos \frac{5}{2}(A-B)=0$.
Let
$C=108^{\circ} \Rightarrow 5 C=540^{\circ}, A+B=72^{\circ} \Rightarrow L=2 \sin \frac{5}{2}(A+B) \times \cos \frac{5}{2}(A-B)=0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,910 |
3.481 Find the minimum value of the expression
$A=\frac{\operatorname{ctg} \alpha-\operatorname{tg} \alpha}{\cos 4 \alpha+1}$ for $0<\alpha<\frac{\pi}{4}$. | ## Solution.
$\operatorname{ctg} \alpha-\operatorname{tg} \alpha=\frac{\cos ^{2} \alpha-\sin ^{2} \alpha}{\sin \alpha \cdot \cos \alpha}=\frac{2 \cos 2 \alpha}{\sin 2 \alpha} \Rightarrow A=\frac{2 \cos 2 \alpha}{\sin 2 \alpha \cdot 2 \cos ^{2} 2 \alpha}=\frac{2}{\sin 4 \alpha}$.
Since $0<\alpha<\frac{\pi}{4}$, the mi... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,911 |
### 3.482 Find the maximum value of the expression
$$
A=\frac{\cos 2 \alpha+1}{\operatorname{ctg} \frac{\alpha}{2}-\operatorname{tg} \frac{\alpha}{2}} \text { for } 0<\alpha<\frac{\pi}{2}
$$ | Solution. This problem is analogous to problem 3.481.
Answer: $\frac{1}{2}$ when $\alpha=\frac{\pi}{4}$. | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,912 |
### 3.485 Find the maximum value of the expression
$$
A=\frac{1}{\sin ^{6} \alpha+\cos ^{6} \alpha} \text { for } 0 \leq \alpha \leq \frac{\pi}{2}
$$ | ## Solution.
$\sin ^{6} \alpha+\cos ^{6} \alpha=\sin ^{4} \alpha-\sin ^{2} \alpha \cdot \cos ^{2} \alpha+\cos ^{4} \alpha=1-3 \sin ^{2} \alpha \cdot \cos ^{2} \alpha=$ $=1-\frac{3}{4} \sin ^{2} 2 \alpha=\frac{1+3 \cos ^{2} 2 \alpha}{4} \Rightarrow A=\frac{4}{1+3 \cos ^{2} 2 \alpha}$.
From this, it is clear that $A$ t... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,913 |
### 3.486 Find the maximum value of the expression
$A=\frac{1}{\sin ^{4} \alpha+\cos ^{4} \alpha}$ for $0 \leq \alpha \leq \frac{\pi}{2}$. | Solution. $\sin ^{4} \alpha+\cos ^{4} \alpha=1-2 \sin ^{2} \alpha \cdot \cos ^{2} \alpha=1-\frac{\sin ^{2} 2 \alpha}{2}=\frac{1+\cos ^{2} 2 \alpha}{2} \Rightarrow$
$\Rightarrow A=\frac{2}{1+\cos ^{2} 2 \alpha}$. From this, it is clear that $A$ takes its maximum value when $1+\cos ^{2} 2 \alpha$ takes its minimum value,... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,914 |
### 3.488 Prove that if the equality
$(1-\sin \alpha)(1-\sin \beta)(1-\sin \gamma)=(1+\sin \alpha)(1+\sin \beta)(1+\sin \gamma)$,
is true, then each part of this equality is equal to $|\cos \alpha \cdot \cos \beta \cdot \cos \gamma|$. | Solution. Multiplying both sides of the equality by $(1+\sin \alpha)(1+\sin \beta)(1+\sin \gamma)$, we get $\left(1-\sin ^{2} \alpha\right)\left(1-\sin ^{2} \beta\right)\left(1-\sin ^{2} \gamma\right)=(1+\sin \alpha)^{2}(1+\sin \beta)^{2}(1+\sin \gamma)^{2} \Leftrightarrow$ $\Leftrightarrow|\cos \alpha \cdot \cos \beta... | |\cos\alpha\cdot\cos\beta\cdot\cos\gamma| | Algebra | proof | Yes | Yes | olympiads | false | 51,916 |
3.489 Prove that for $0<\varphi<\frac{\pi}{4}$ the equality holds
$$
1-\operatorname{tg} \varphi+\operatorname{tg}^{2} \varphi-\operatorname{tg}^{3} \varphi+\ldots=\frac{\sqrt{2} \cos \varphi}{2 \sin \left(\frac{\pi}{4}+\varphi\right)}
$$ | Solution.
$$
1-\operatorname{tg} \varphi+\operatorname{tg}^{2} \varphi-\operatorname{tg}^{3} \varphi+\ldots=\frac{1}{1+\operatorname{tg} \varphi}=\frac{\cos \varphi}{\sin \varphi+\cos \varphi}=\frac{\cos \varphi}{\sqrt{2} \sin \left(\varphi+\frac{\pi}{4}\right)}=\frac{\sqrt{2} \cos \varphi}{2 \sin \left(\frac{\pi}{4}+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,917 |
### 3.490 Find the minimum value of the expression
$A=\sin ^{6} \alpha+\cos ^{6} \alpha$ for $0 \leq \alpha \leq \frac{\pi}{2}$. | Solution. This problem is identical to problem 3.485.
Answer: $\frac{1}{4}$ when $\alpha=\frac{\pi}{4}$. | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,918 |
### 3.491 Find the minimum value of the expression
$A=\sin ^{4} \alpha+\cos ^{4} \alpha$ for $0 \leq \alpha \leq \frac{\pi}{2}$ | Solution. This problem is identical to problem 3.486.
Answer: $\frac{1}{2}$ when $\alpha=\frac{\pi}{4}$. | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,919 |
3.492 Prove that if $\alpha$ is constant, then the function
$f(x)=\cos ^{2} x+\cos ^{2}(x+\alpha)-2 \cos \alpha \cdot \cos x \cdot \cos (x+\alpha)$ is constant. | Solution.
$$
\begin{aligned}
& f(x)=\cos ^{2} x+\cos ^{2}(x+\alpha)-\cos (\alpha+x) \cdot(\cos (\alpha+x)+\cos (\alpha-x))= \\
& =\cos ^{2} x-\cos (\alpha+x) \cdot \cos (\alpha-x)=\frac{1+\cos 2 x}{2}-\frac{\cos 2 x+\cos 2 \alpha}{2}=\frac{1-\cos 2 \alpha}{2}
\end{aligned}
$$
That is, $f(x)$ does not depend on $x$, a... | \frac{1-\cos2\alpha}{2} | Algebra | proof | Yes | Yes | olympiads | false | 51,920 |
### 3.493 Find the sum $1+\cos 4 \alpha+\cos 8 \alpha+\ldots+\cos 4 n \alpha$. | ## Solution.
$2 \sin 2 \alpha(1+\cos 4 \alpha+\cos 8 \alpha+\ldots+\cos 4 n \alpha)=2 \sin 2 \alpha+2 \sin 2 \alpha \cdot \cos 4 \alpha+\ldots+$ $+2 \sin 2 \alpha \cdot \cos 4 n \alpha=2 \sin 2 \alpha+(\sin 6 \alpha-\sin 2 \alpha)+(\sin 10 \alpha-\sin 6 \alpha)+\ldots+$ $+(\sin (4 n \alpha-2 \alpha)-\sin (4 n \alpha-6... | \frac{\sin2\alpha(n+1)\cdot\cos2n\alpha}{\sin} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,921 |
3.494 Prove the identity: $\operatorname{tg} 5^{\circ} \cdot \operatorname{tg} 20^{\circ}+\operatorname{tg} 5^{\circ} \cdot \operatorname{tg} 65^{\circ}+\operatorname{tg} 20^{\circ} \cdot \operatorname{tg} 65^{\circ}=1$. | Solution. $\operatorname{tg} 5^{\circ} \cdot\left(\operatorname{tg} 20^{\circ}+\operatorname{tg} 65^{\circ}\right)=1-\operatorname{tg} 20^{\circ} \cdot \operatorname{tg} 65^{\circ} \Leftrightarrow \operatorname{tg} 5^{\circ} \cdot \frac{\operatorname{tg} 20^{\circ}+\operatorname{tg} 65^{\circ}}{1-\operatorname{tg} 20^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,922 |
3.496 Let $A, B, C$ be the angles of a triangle. Prove that $8 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \leq 1$. | Solution. The original inequality is equivalent to the following:
\[
\begin{aligned}
& 8\left(\sin \frac{A}{2} \cdot \sin \frac{B}{2}\right) \cdot \cos \frac{A+B}{2} \leq 1 \Leftrightarrow 4 \cos \frac{A-B}{2} \cdot \cos \frac{A+B}{2}-4 \cos ^{2} \frac{A+B}{2} \leq 1 \Leftrightarrow \\
& \Leftrightarrow 4 \cos ^{2} \f... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 51,924 |
3.498 Given: $\operatorname{arctg} x+\operatorname{arctg} y+\operatorname{arctg} z=\frac{\pi}{2}$.
Prove that $x y+y z+x z=1$. | Solution. Let $\operatorname{arctg} z=0 \Rightarrow z=0, \operatorname{arctg} x=\frac{\pi}{2}-\operatorname{arctg} y \Rightarrow$ $\Rightarrow \operatorname{tg} \operatorname{arctg} x=\operatorname{tg}\left(\frac{\pi}{2}-\operatorname{arctg} y\right) \Rightarrow x=\operatorname{ctg} \operatorname{arctg} y \Rightarrow x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,925 |
### 3.499 Let $A, B, C$ be the angles of a triangle. Prove that
$\mathbf{8} \cos A \cdot \cos B \cdot \cos C \leq 1$. | Solution. The original inequality is equivalent to the following:
$-8(\cos A \cdot \cos B) \cdot \cos (A+B) \leq 1 \Leftrightarrow$
$\Leftrightarrow-4(\cos (A-B)+\cos (A+B)) \cdot \cos (A+B) \leq \cos ^{2}(A-B)+\sin ^{2}(A-B) \Leftrightarrow$,
$\Leftrightarrow(2 \cos (A+B)+\cos (A-B))^{2}+\sin ^{2}(A-B) \geq 0$ whic... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 51,926 |
### 3.500 Let A, B, C be the angles of a triangle. Prove that
$\sin ^{2} A+\sin ^{2} B+\sin ^{2} C \leq \frac{9}{4}$. | ## Solution.
$$
\begin{aligned}
& 1-\cos 2 A+1-\cos 2 B+1-\cos 2 C \leq \frac{9}{2} \Leftrightarrow \frac{3}{2}+\cos 2 A+\cos 2 B+\cos 2 C \geq 0 \Leftrightarrow \\
& \Leftrightarrow \frac{3}{2}+2 \cos (A+B) \cdot \cos (A-B)+\cos 2(A+B) \geq 0 \Leftrightarrow \\
& \Leftrightarrow 1+4 \cos (A+B) \cdot \cos (A-B)+4 \cos... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 51,927 |
$8.407 \frac{1+\tan x+\tan^{2} x+\ldots+\tan^{n} x+\ldots}{1-\tan x+\tan^{2} x-\ldots(-1)^{n} \tan^{n} x+\ldots}=1+\sin 2 x,|\tan x|<1$ | Solution. $1+\operatorname{tg} x+\operatorname{tg}^{2} x+\ldots+\operatorname{tg}^{n} x+\ldots=\frac{1}{1-\operatorname{tg} x}, 1-\operatorname{tg} x+\operatorname{tg}^{2} x-\ldots+(-1)^{n} \operatorname{tg}^{n} x+\ldots=\frac{1}{1+\operatorname{tg} x}$
Therefore, the original equation is equivalent to the following:
... | \pik | Algebra | proof | Yes | Yes | olympiads | false | 51,928 |
$8.408 \operatorname{tg} x-\sin 2 x-\cos 2 x \cdot\left(1-\frac{2}{\cos x}\right)=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$8.408 \tan x - \sin 2x - \cos 2x \cdot \left(1 - \frac{2}{\cos x}\right) = 0$. | Solution. $\frac{\sin x}{\cos x}-\sin 2 x-\cos 2 x \cdot \frac{\cos x-2}{\cos x}=0 \Leftrightarrow$ 142
$\Leftrightarrow\left\{\begin{array}{l}\sin x-\sin 2 x \cdot \cos x-\cos 2 x \cdot(\cos x-2)=0, \\ \cos x \neq 0\end{array} \Leftrightarrow\right.$
$\Leftrightarrow\left\{\begin{array}{l}\sin x \cdot\left(1-2 \cos ^... | \frac{\pi}{4}(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,929 |
$8.409 \frac{1-\sin x+\sin ^{2} x-\ldots+(-1)^{n} \sin ^{n} x+\ldots}{1+\sin x+\sin ^{2} x+\ldots+\sin ^{n} x+\ldots}=\frac{1-\cos 2 x}{1+\cos 2 x},|\sin x|<1$.
$8.409 \frac{1-\sin x+\sin ^{2} x-\ldots+(-1)^{n} \sin ^{n} x+\ldots}{1+\sin x+\sin ^{2} x+\ldots+\sin ^{n} x+\ldots}=\frac{1-\cos 2 x}{1+\cos 2 x},|\sin x|<1... | Solution. The original equation is equivalent to the following:
\[
\begin{aligned}
& \left\{\begin{array}{l}
\frac{1-\sin x}{1+\sin x}=\frac{1-\cos 2 x}{1+\cos 2 x}, \\
|\sin x|<1
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
\frac{1-\sin x}{1+\sin x}=\frac{\sin ^{2} x}{\cos ^{2} x}, \\
|\sin x|<1
\en... | (-1)^{k}\frac{\pi}{6}+\pik | Algebra | proof | Yes | Yes | olympiads | false | 51,930 |
$8.410 \sqrt{3} \sin t-\sqrt{2 \sin ^{2} t-\sin 2 t+3 \cos ^{2} t}=0$. | ## Solution.
$$
\begin{aligned}
& \sqrt{3} \sin t=\sqrt{2 \sin ^{2} t-\sin 2 t+3 \cos ^{2} t} \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l }
{ \sin t \geq 0 , } \\
{ 3 \sin ^{2} t = 2 \sin ^{2} t - \sin 2 t + 3 \cos ^{2} t }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\sin t \geq 0, \\
\sin ^{... | t_{1}=\frac{\pi}{4}+2k\pi,t_{2}=-\arctan3+\pi(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,931 |
$8.413 \cos x \cdot \sqrt{\tan^{2} x-\sin ^{2} x}+\sin x \cdot \sqrt{\cot^{2} x-\cos ^{2} x}=2 \sin x$. | Solution. The original equation is equivalent to the following:
$\cos x \cdot \sqrt{\sin ^{2} x \cdot\left(\frac{1}{\cos ^{2} x}-1\right)}+\sin x \cdot \sqrt{\cos ^{2} x \cdot\left(\frac{1}{\sin ^{2} x}-1\right)}=2 \sin x \Leftrightarrow$
$\Leftrightarrow \frac{\cos x \cdot \sin ^{2} x}{|\cos x|}+\frac{\sin x \cdot \c... | x_{1}=\frac{\pi}{6}+\pik,x_{2}=\arcsin\frac{1-\sqrt{3}}{2}+\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,932 |
$8.414 \cos x+\sqrt{\frac{3}{2}-\cos ^{2} x}-\cos x \cdot \sqrt{\frac{3}{2}-\cos ^{2} x}=1$. | Solution. The original equation is equivalent to the following:
$(1-\cos x) \cdot\left(\sqrt{\frac{3}{2}-\cos ^{2} x}-1\right)=0 \Leftrightarrow\left[\begin{array}{l}\cos x=1, \\ \frac{3}{2}-\cos ^{2} x=1\end{array} \Leftrightarrow\left[\begin{array}{l}x=2 \pi k, \\ x=\frac{\pi}{4}+\frac{\pi k}{2}\end{array}\right.\ri... | x_{1}=2\pik,x_{2}=\frac{\pi}{4}(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,933 |
$8.415 \sqrt[5]{\frac{1}{2}-\sin x}+\sqrt[5]{\frac{1}{2}+\sin x}=1$ | Solution. Let $\quad u=\sqrt[5]{\frac{1}{2}-\sin x}, v=\sqrt[5]{\frac{1}{2}+\sin x} . \quad$ Then $\left\{\begin{array}{l}u+v=1, \\ u^{5}+v^{5}=1\end{array}\right.$ $u^{5}+v^{5}=(u+v)\left(u^{4}-u^{3} v+u^{2} v^{2}-u v^{3}+v^{4}\right)=u^{4}+2 u^{2} v^{2}+v^{4}-u v\left(u^{2}+v^{2}\right)-u^{2} v^{2}=$ $=\left(u^{2}+v^... | \frac{\pi}{6}(6k\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,934 |
$8.416 \sqrt{1+\sin x}-\sqrt{1-\sin x}=1+\cos x$. | ## Solution.
$\left\{\begin{array}{l}\sqrt{1+\sin x} \geq \sqrt{1-\sin x}, \\ (\sqrt{1+\sin x}-\sqrt{1-\sin x})^{2}=(1+\cos x)^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin x \geq 0, \\ 2-2|\cos x|=1+2 \cos x+\cos ^{2} x\end{array} \Leftrightarrow\right.\right.$
$\Leftrightarrow\left\{\begin{array}{l}\sin... | x_{1}=\arccos(\sqrt{5}-2)+2\pik,x_{2}=\pi(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,935 |
$8.417 \sqrt{1+3 \cot x}+\sqrt{\frac{\tan x}{3+\tan x}}=\frac{5}{2}$. | ## Solution.
$\sqrt{\frac{3+\operatorname{tg} x}{\operatorname{tg} x}}+\sqrt{\frac{\operatorname{tg} x}{3+\operatorname{tg} x}}=\frac{5}{2}$. Let $y=\sqrt{\frac{\operatorname{tg} x}{3+\operatorname{tg} x}}$, then
$2 y^{2}-5 y+2=0 \Leftrightarrow\left[\begin{array}{l}y=2, \\ y=\frac{1}{2}\end{array} \Leftrightarrow\le... | x_{1}=-\operatorname{arctg}4+\pik,x_{2}=\frac{\pi}{4}+\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,936 |
$8.418 \sqrt[4]{\frac{1}{2}-\cos 2 x}+\sqrt[4]{\frac{1}{2}+\cos 2 x}=1$ | ## Solution.
Let $u=\sqrt[4]{\frac{1}{2}-\cos 2 x}, v=\sqrt[4]{\frac{1}{2}+\cos 2 x}, \quad$ then $\left\{\begin{array}{l}u+v=1, \\ u^{4}+v^{4}=1\end{array} \Leftrightarrow\right.$
$\Leftrightarrow\left\{\begin{array}{l}u+v=1, \\ \left((u+v)^{2}-2 u v\right)^{2}-2 u^{2} v^{2}=1\end{array} \Leftrightarrow\left\{\begin... | \frac{\pi}{6}(3k\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,937 |
$8.419 \sin x+\sqrt{2-\sin ^{2} x}+\sin x \cdot \sqrt{2-\sin ^{2} x}=3$. | Solution. Let $u=\sin x, v=\sqrt{2-\sin ^{2} x}$.
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ u + v + u v = 3 , } \\
{ u ^ { 2 } + v ^ { 2 } = 2 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ u + v = 3 - u v , } \\
{ ( u + v ) ^ { 2 } = 2 + 2 u v }
\end{array} \Leftrightarrow \left\{\begin{array}{l... | \frac{\pi}{6}(6k\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,938 |
$8.423 \sqrt[3]{\sin ^{2} x}-\sqrt[3]{\cos ^{2} x}=\sqrt[3]{2 \cos 2 x}$. | Solution. The original equation is equivalent to the following:
$$
\left(\sqrt[3]{\sin ^{2} x}-\sqrt[3]{\cos ^{2} x}\right)^{3}=2\left(\cos ^{2} x-\sin ^{2} x\right)
$$
Let: $u=\sqrt[3]{\sin ^{2} x}, v=\sqrt[3]{\cos ^{2} x}$. Then
$$
(u-v)^{3}=2\left(v^{3}-u^{3}\right) \Leftrightarrow(u-v)\left(u^{2}+v^{2}\right)=0 ... | \frac{\pi}{4}(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,939 |
$8.424 \cos x+\sqrt{\sin ^{2} x-2 \sin 2 x+4 \cos ^{2} x}=0$. | ## Solution.
$\sqrt{\sin ^{2} x-4 \sin x \cdot \cos x+4 \cos ^{2} x}=-\cos x \Leftrightarrow|2 \cos x-\sin x|=-\cos x \Leftrightarrow$
1) $\left\{\begin{array}{l}\cos x \leq 0, \\ 2 \cos x-\sin x=-\cos x\end{array} \Leftrightarrow\left\{\begin{array}{l}\cos x \leq 0, \\ \operatorname{tg} x=3\end{array} \Leftrightarro... | x_{1}=\operatorname{arctg}3+\pi(2k+1),\quadx_{2}=\frac{\pi}{4}+\pi(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,940 |
$8.425 \sqrt{\cos 2 x}+\sqrt{1+\sin 2 x}=2 \sqrt{\sin x+\cos x}$. | Solution. The original equation is equivalent to the following:
$\sqrt{(\cos x-\sin x)(\cos x+\sin x)}+\sqrt{(\sin x+\cos x)^{2}}=2 \sqrt{\sin x+\cos x}$.
1) $\sin x+\cos x=0 \Leftrightarrow x=-\frac{\pi}{4}+\pi k$.
2) $\left\{\begin{array}{l}\sin x+\cos x>0 \\ \sqrt{\cos x-\sin x}+\sqrt{\cos x+\sin x}=2 .\end{array}... | x_{1}=\frac{\pi}{4}(4k-1),x_{2}=2\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,941 |
8.426 \frac{1-2 \cos ^{2} x}{\sin x \cdot \cos x}+2 \tan 2 x+\cot ^{3} 4 x=3
$$ | Solution. Transform the original equation:
$$
\begin{aligned}
& -\frac{2 \cos 2 x}{\sin 2 x}+\frac{2}{\operatorname{ctg} 2 x}+\operatorname{ctg}^{3} 4 x=3 \Leftrightarrow \frac{2\left(1-\operatorname{ctg}^{2} 2 x\right)}{\operatorname{ctg} 2 x}+\operatorname{ctg}^{3} 4 x=3 \Leftrightarrow \\
& \Leftrightarrow-4 \opera... | x_{1}=\frac{\pi}{16}(4k+3),x_{2}=\frac{1}{4}(\operatorname{arcctg}\frac{1\\sqrt{13}}{2}+\pik) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,942 |
$8.427 \sqrt[4]{10+8 \sin ^{2} x}-\sqrt[4]{8 \cos ^{2} x-1}=1$. | ## Solution.
Let $\sqrt[4]{10+8 \sin ^{2} x}=u, \sqrt[4]{8 \cos ^{2} x-1}=v$.
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ u - v = 1 , } \\
{ u ^ { 4 } + v ^ { 4 } = 1 7 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ u - v = 1 , } \\
{ ( u ^ { 2 } - v ^ { 2 } ) ^ { 2 } + 2 u ^ { 2 } v ^ { 2 } = 1 7... | \frac{\pi}{3}(3k-\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,943 |
$8.430 \sin \pi \sqrt{t}+\sin \pi t=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$8.430 \sin \pi \sqrt{t}+\sin \pi t=0$. | Solution. $\sin \frac{\pi(\sqrt{t}+t)}{2} \cdot \cos \frac{\pi(t-\sqrt{t})}{2}=0 \Leftrightarrow\left[\begin{array}{l}\pi(t+\sqrt{t})=2 \pi n, \\ \pi(t-\sqrt{t})=\pi+2 \pi k\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left[\begin{array}{l}t+\sqrt{t}-2 n=0, n \in Z, \\ t-\sqrt{t}-(2 k+1)=0, k \in Z\end{array} \L... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,945 |
$8.433 \cos ^{-4} x+\cos ^{4} x=1+\cos 2 x-2 \sin ^{2} 2 x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$8.433 \cos ^{-4} x+\cos ^{4} x=1+\cos 2 x-2 \sin ^{2} 2 x$. | Solution.
$\frac{1}{\cos ^{4} x}+\cos ^{4} x=2 \cos ^{2} x-2 \sin ^{2} 2 x \Leftrightarrow\left(\frac{1}{\cos ^{2} x}-\cos ^{2} x\right)^{2}=-2\left(\sin ^{2} x+\sin ^{2} 2 x\right)$, but $\left(\frac{1}{\cos ^{2} x}-\cos ^{2} x\right)^{2} \geq 0,-2\left(\sin ^{2} 2 x+\sin ^{2} x\right) \leq 0$, from which $\frac{1}{\... | \pik | Algebra | proof | Yes | Yes | olympiads | false | 51,946 |
$8.434 \cos ^{-4} x+8 \cos ^{-1} x-7=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$8.434 \cos ^{-4} x+8 \cos ^{-1} x-7=0$. | Solution. Let $y=\frac{1}{\cos x}$, then $y^{4}+8 y-7=0 \Leftrightarrow$
$$
\begin{aligned}
& y^{4}+2 y^{2}+1-2 y^{2}+8 y-8=0 \Leftrightarrow\left(y^{2}+1\right)^{2}-2(y-2)^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(y^{2}+y \sqrt{2}-2 \sqrt{2}+1\right)\left(y^{2}-y \sqrt{2}+2 \sqrt{2}+1\right)=0 \Leftrightarrow y... | \\arccos\frac{-2}{\sqrt{2}+\sqrt{8\sqrt{2}-2}}+2\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,947 |
$8.435 \sin ^{10} 3 x+\cos ^{10} 3 x=4 \frac{\sin ^{6} 3 x+\cos ^{6} 3 x}{4 \cos ^{2} 6 x+\sin ^{2} 6 x}$. | Solution. Transform the right side of the original equation:
$4 \frac{\sin ^{6} 3 x+\cos ^{6} 3 x}{4 \cos ^{2} 6 x+\sin ^{2} 6 x}=\frac{4\left(\frac{(1-\cos 6 x)^{3}}{2^{3}}+\frac{(1+\cos 6 x)^{3}}{2^{3}}\right)}{1+3 \cos ^{2} 6 x}=\frac{1}{2} \cdot \frac{2+6 \cos ^{2} 6 x}{3 \cos ^{2} 6 x+1}=1$.
From here $\sin ^{10... | \frac{\pik}{6},\frac{\pi}{8}(2k+1),\pik,\frac{\pi}{6}(4n+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,948 |
$8.438 \cos \left(x-\frac{\pi}{4}\right) \cdot\left(1-4 \cos ^{2} 2 x\right)-2 \cos 4 x=3$. | ## Solution.
$$
\begin{aligned}
& \cos \left(x-\frac{\pi}{4}\right) \cdot\left(1-4 \cos ^{2} 2 x\right)=1+2(\cos 4 x+1) \Leftrightarrow \\
& \Leftrightarrow \cos \left(x-\frac{\pi}{4}\right) \cdot\left(1-4 \cos ^{2} 2 x\right)=1+4 \cos ^{2} 2 x \Leftrightarrow \cos \left(x-\frac{\pi}{4}\right)=\frac{1+4 \cos ^{2} 2 x}... | \frac{\pi}{4}(8k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,949 |
$8.43918 \cos ^{2} x+5\left(3 \cos x+\frac{1}{\cos x}\right)+\frac{2}{\cos ^{2} x}+5=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$8.43918 \cos ^{2} x+5\left(3 \cos x+\frac{1}{\cos x}\right)+\frac{2}{\cos ^{2} x}+5=0$. | Solution. Rewrite the equation as follows:
$$
\begin{aligned}
& 2\left(9 \cos ^{2} x+\frac{1}{\cos ^{2} x}\right)+5\left(3 \cos x+\frac{1}{\cos x}\right)+5=0 \\
& \text { Let } \quad y=3 \cos x+\frac{1}{\cos x}, \text { then } 2\left(y^{2}-6\right)+5 y+5=0 \Leftrightarrow \\
& \Leftrightarrow 2 y^{2}+5 y-7=0 \Leftrigh... | x_{1}=\\frac{2\pi}{3}+2\pik,x_{2}=\\arccos(-\frac{2}{3})+2\pik | Algebra | proof | Yes | Yes | olympiads | false | 51,950 |
$8.440 \operatorname{tg}(\pi \cdot \operatorname{ctg} x)=\operatorname{ctg}(\pi \cdot \operatorname{tg} x)$
$8.440 \tan(\pi \cdot \cot x)=\cot(\pi \cdot \tan x)$ | Solution.
If $\operatorname{tg} x=l$, then $l+\frac{1}{l}=\frac{1}{2}+k \Rightarrow \frac{1}{2}-\frac{1}{l}$ is an integer, which is only possible when $l= \pm 2$, i.e., $k=2$ or $k=-3$. Th... | \frac{(-1)^{n}}{2}\arcsin\frac{4}{1+2k}+\frac{\pin}{2},[\begin{pmatrix}k=3,4,\ldots\\k=-4,-5,\ldots\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,951 |
$8.441 \cos ^{-4} x-2 \cos ^{-2} x-12 \tan x-16=0$. | Solution. The equation is equivalent to the following:
$\left(1+\operatorname{tg}^{2} x\right)^{2}-2\left(1+\operatorname{tg}^{2} x\right)-12 \operatorname{tg} x-16=0 \Leftrightarrow$
$\Leftrightarrow\left(1+\operatorname{tg}^{2} x\right)^{2}-\left(2 \operatorname{tg}^{2} x+12 \operatorname{tg} x+18\right)=0 \Leftrig... | \operatorname{arctg}\frac{1\\sqrt{6\sqrt{2}-1}}{\sqrt{2}}+\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,952 |
$8.442 \sin ^{8} 2 x+\cos ^{8} 2 x=\frac{41}{128}$.
The translation is as follows:
$8.442 \sin ^{8} 2 x+\cos ^{8} 2 x=\frac{41}{128}$. | ## Solution.
$\left(\sin ^{4} 2 x-\cos ^{4} 2 x\right)^{2}+2 \cos ^{4} 2 x \cdot \sin ^{4} 2 x=\frac{41}{128} \Leftrightarrow 8 \cos ^{2} 4 x+\sin ^{4} 4 x=\frac{41}{16} \Leftrightarrow$ $\Leftrightarrow \sin ^{4} 4 x-8 \sin ^{2} 4 x+\frac{87}{16}=0 \Leftrightarrow \sin ^{2} 4 x=4 \pm \frac{13}{4} \Leftrightarrow \sin... | \frac{\pi}{12}(3k\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,953 |
$8.444 \frac{\tan x}{2-\frac{1}{\cos ^{2} x}} \cdot(\sin 3 x-\sin x)=\frac{2}{\cot^{2} x-3}$. | Solution. The original equation is equivalent to the following system:
$$
\left\{\begin{array}{l}
\frac{\sin x \cdot \cos x}{2 \cos ^{2} x-1} \cdot 2 \sin x \cdot \cos 2 x=\frac{2 \sin ^{2} x}{\cos ^{2} x-3 \sin ^{2} x}, \\
\cos x \neq 0, \\
\sin x \neq 0
\end{array} \Leftrightarrow\right.
$$
$$
\begin{aligned}
& \Le... | \\frac{2\pi}{3}+2\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,954 |
$8.446 \tan x + \cot x - \cos 4x = 3$. | Solution.
$\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=2+(1+\cos 4 x) \Leftrightarrow \frac{2}{\sin 2 x}=2+2 \cos ^{2} 2 x \Leftrightarrow$
$\Leftrightarrow\left\{\begin{array}{l}\sin 2 x \neq 0, \\ \sin 2 x+\cos ^{2} 2 x \cdot \sin 2 x=1\end{array} \Leftrightarrow \sin ^{3} 2 x-2 \sin 2 x+1=0 \Leftrightarrow\right.$
... | x_{1}=\frac{\pi}{4}(4k+1),x_{2}=\frac{1}{2}(-1)^{k}\arcsin\frac{\sqrt{5}-1}{2}+\frac{\pik}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,955 |
$8.447 \tan 2 x \cdot \tan^{2} 3 x \cdot \tan^{2} 5 x=\tan 2 x+\tan^{2} 3 x-\tan^{2} 5 x$. | ## Solution.
$\operatorname{tg} 2 x \cdot\left(1-\operatorname{tg}^{2} 3 x \cdot \operatorname{tg}^{2} 5 x\right)=\operatorname{tg}^{2} 5 x-\operatorname{tg}^{2} 3 x \Leftrightarrow$
$\Leftrightarrow \frac{\sin 2 x}{\cos 2 x} \cdot\left(1-\frac{\sin 3 x \cdot \sin 5 x}{\cos 3 x \cdot \cos 5 x}\right)\left(1+\frac{\si... | x_{1}=\pik,x_{2}=\frac{\pi}{32}(4k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,956 |
$8.450 \cos ^{4} x+4 \cos x-1=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$8.450 \cos ^{4} x+4 \cos x-1=0$. | Solution. $\cos ^{4} x+2 \cos ^{2} x+1-2 \cos ^{2} x+4 \cos x-2=0 \Leftrightarrow$
$\Leftrightarrow\left(\cos ^{2} x+1\right)^{2}-2(\cos x-1)^{2}=0 \Leftrightarrow$
$\Leftrightarrow\left(\cos ^{2} x+\sqrt{2} \cos x-\sqrt{2}+1\right)\left(\cos ^{2} x-\sqrt{2} \cos x+\sqrt{2}+1\right)=0 \Leftrightarrow$
$\Leftrightarro... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,957 | |
$8.451 \frac{1-\cos 2 x+\cos ^{2} 2 x-\ldots+(-1)^{n} \cos ^{n} 2 x+\ldots}{1+\cos 2 x+\cos ^{2} 2 x+\ldots+\cos ^{n} 2 x+\ldots}=\frac{1}{3} \tan^{4} x,|\cos 2 x| \neq 1$. | Solution. The original equation is equivalent to the following:
\[
\begin{aligned}
& \frac{1-\cos 2 x}{1+\cos 2 x}=\frac{1}{3} \tan^{4} x \Leftrightarrow \frac{1-\cos 2 x}{1+\cos 2 x}=\frac{(1-\cos 2 x)^{2}}{3(1+\cos 2 x)^{2}} \Leftrightarrow 3=\frac{1-\cos 2 x}{1+\cos 2 x} \Leftrightarrow \\
& \Leftrightarrow \cos 2 ... | x_{1}=-\arctan\frac{3}{2}+\pik,x_{2}=\frac{\pi}{4}\\arccos\frac{\sqrt{2}-2}{2}+2\pik | Algebra | proof | Yes | Yes | olympiads | false | 51,958 |
$8.453 \tan^{3} x+\tan^{2} x+\cot^{2} x+\cot^{3} x-4=0$. | ## Solution.
$\left(\operatorname{tg}^{3} x+\operatorname{ctg}^{3} x\right)+\left(\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x+2\right)-6=0 \Leftrightarrow$
$\Leftrightarrow(\operatorname{tg} x+\operatorname{ctg} x)\left(\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x-1\right)+(\operatorname{tg} x+\operatorname{... | \frac{\pi}{4}(4k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,959 |
$8.454 \cos \sqrt{x}=\cos x$. | ## Solution.
$\cos x-\cos \sqrt{x}=0 \Leftrightarrow \sin \frac{x-\sqrt{x}}{2} \cdot \sin \frac{x+\sqrt{x}}{2}=0 \Leftrightarrow\left[\begin{array}{l}x-\sqrt{x}-2 \pi k=0, \\ x+\sqrt{x}-2 \pi k=0\end{array} \Leftrightarrow\right.$
$\Leftrightarrow\left[\begin{array}{l}\sqrt{x}=\frac{1 \pm \sqrt{1+8 \pi k}}{2}, \\ \sq... | \frac{1\\sqrt{1+8\pik}}{2}+2\pik,k=0,1,\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,960 |
$8.456 \tan^{2} x \cdot \tan^{2} 3 x \cdot \tan 4 x=\tan^{2} x-\tan^{2} 3 x+\tan 4 x$. | ## Solution.
$\operatorname{tg} 4 x \cdot\left(1-\operatorname{tg}^{2} x \cdot \operatorname{tg}^{2} 3 x\right)=(\operatorname{tg} 3 x-\operatorname{tg} x)(\operatorname{tg} 3 x+\operatorname{tg} x) \Leftrightarrow$ $\Leftrightarrow \frac{\sin 4 x}{\cos 4 x} \cdot\left(1-\frac{\sin x \cdot \sin 3 x}{\cos x \cdot \cos ... | x_{1}=\pik,x_{2}=\frac{\pi}{4}(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,961 |
$8.457 \cos 6 x+\sin \frac{5 x}{2}=2$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$8.457 \cos 6 x+\sin \frac{5 x}{2}=2$. | Solution. The equation is equivalent to the system:
$\left\{\begin{array}{l}\cos 6 x=1, \\ \sin \frac{5 x}{2}=1\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{\pi l}{3}, l \in Z, \\ x=\frac{\pi}{5}+\frac{4 \pi n}{5}, n \in Z\end{array} \Leftrightarrow \frac{\pi l}{3}=\frac{\pi}{5}+\frac{4 \pi n}{5} \Rightarro... | \pi(1+4k) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,962 |
$8.459 \cos ^{2} x^{2}\left(\operatorname{tg} x^{2}+2 \operatorname{tg} x\right)+\operatorname{tg}^{3} x \cdot\left(1-\sin ^{2} x^{2}\right)\left(2-\operatorname{tg} x \cdot \operatorname{tg} x^{2}\right)=0$.
$8.459 \cos ^{2} x^{2}\left(\tan x^{2}+2 \tan x\right)+\tan^{3} x \cdot\left(1-\sin ^{2} x^{2}\right)\left(2-\... | Solution. The original equation is equivalent to
$\operatorname{tg} x^{2}+2 \operatorname{tg} x+2 \operatorname{tg}^{3} x-\operatorname{tg}^{4} x \cdot \operatorname{tg} x^{2}=0 \Leftrightarrow \operatorname{tg}^{2}\left(1-\operatorname{tg}^{4} x\right)+2 \operatorname{tg} x \cdot\left(1+\operatorname{tg}^{2} x\right)... | -1\\sqrt{\pik+1},k=0,1,2,\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,963 |
$8.460 \frac{1-\tan x+\tan^{2} x-\ldots+(-1)^{n} \cdot \tan^{n} x+\ldots}{1+\tan x+\tan^{2} x+\ldots+\tan^{n} x+\ldots}=1+\sin 2 x,|\tan x|<1$. | Solution. The original equation is equivalent to the following:
\[
\begin{aligned}
& \frac{1-\tan x}{1+\tan x}=1+\sin 2 x \Leftrightarrow \frac{1-\tan x}{1+\tan x}=1+\frac{2 \tan x}{1+\tan^{2} x} \Leftrightarrow \\
& \Leftrightarrow(1-\tan x)\left(1+\tan^{2} x\right)=(1+\tan x)^{3} \Leftrightarrow \tan x \cdot\left(\t... | \pik | Algebra | proof | Yes | Yes | olympiads | false | 51,964 |
$8.462 \operatorname{tg}^{2} x \cdot \operatorname{ctg}^{2} 2 x \cdot \operatorname{ctg} 3 x=\operatorname{tg}^{2} x-\operatorname{ctg}^{2} 2 x+\operatorname{ctg} 3 x$.
$8.462 \tan^{2} x \cdot \cot^{2} 2 x \cdot \cot 3 x=\tan^{2} x-\cot^{2} 2 x+\cot 3 x$. | Solution. $\operatorname{ctg} 3 x \cdot\left(1-\operatorname{tg}^{2} x \cdot \operatorname{ctg}^{2} 2 x\right)=\operatorname{ctg}^{2} 2 x-\operatorname{tg}^{2} x \Leftrightarrow$
$$
\Leftrightarrow \frac{\cos 3 x}{\sin 3 x}\left(1-\frac{\sin x \cdot \cos 2 x}{\cos x \cdot \sin 2 x}\right)\left(1+\frac{\sin x \cdot \co... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,965 |
$8.469 \sin ^{4} x+2 \cos ^{3} x+2 \sin ^{2} x-\cos x+1=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$8.469 \sin ^{4} x+2 \cos ^{3} x+2 \sin ^{2} x-\cos x+1=0$. | Solution. $\sin ^{4} x+2 \sin ^{2} x+1=\cos x \cdot\left(1-2 \cos ^{2} x\right) \Leftrightarrow\left(\sin ^{2} x+1\right)^{2}=-\cos x \cdot \cos 2 x$.
For any $x\left(\sin ^{2} x+1\right)^{2} \geq 1,-\cos x \cdot \cos 2 x \leq 1$. Therefore $\left\{\begin{array}{l}\sin ^{2} x+1=1, \\ -\cos x \cdot \cos 2 x=1\end{array... | \pi(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,967 |
$8.471 \frac{3 \tan x-\tan^{3} x}{2-\frac{1}{\cos ^{2} x}}=\frac{4+2 \cos \frac{6 x}{5}}{\cos 3 x+\cos x}$. | Solution. $\frac{\left(\frac{3 \sin x}{\cos x}-\frac{\sin ^{3} x}{\cos ^{3} x}\right) \cos ^{2} x}{2 \cos ^{2} x-1}=\frac{4+2 \cos \frac{6 x}{5}}{2 \cos 2 x \cdot \cos x} \Leftrightarrow$
$$
\begin{aligned}
& \Leftrightarrow \frac{\sin x \cdot\left(3 \cos ^{2} x-\sin ^{2} x\right)}{\cos x \cdot \cos 2 x}=\frac{2+\cos ... | \frac{5}{6}(\pi+4\pik),k\neq2+3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,968 |
$8.472 \frac{12}{\cos ^{2} x}+\frac{1}{3} \operatorname{ctg}^{2} x+10\left(2 \operatorname{tg} x+\frac{\operatorname{ctg} x}{3}\right)=1$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$8.472 \frac{12}{\cos ^{2} x}+\frac{1}{3} \c... | Solution. $12\left(1+\operatorname{tg}^{2} x\right)+10\left(2 \operatorname{tg} x+\frac{\operatorname{ctg} x}{3}\right)+\frac{1}{3} \operatorname{ctg}^{2} x=1 \Leftrightarrow$.
$\Leftrightarrow 11+3\left(4 \operatorname{tg}^{2} x+\frac{1}{9} \operatorname{ctg}^{2} x\right)+10\left(2 \operatorname{tg} x+\frac{1}{3} \op... | x_{1}=\frac{\pi}{4}(4k-1),x_{2}=-\operatorname{arctg}\frac{1}{6}+\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,969 |
$8.473 \sin ^{5} x+\cos ^{5} x=2-\sin ^{4} x$ | Solution. $\sin ^{5} x \leq \sin ^{2} x, \cos ^{5} x \leq \cos ^{2} x \Rightarrow \sin ^{5} x+\cos ^{5} x \leq 1$. But $2-\sin ^{4} x \geq 1$ for all $x$. Therefore, the original equation is equivalent to $\left\{\begin{array}{l}\sin ^{5} x=\sin ^{2} x \\ \cos ^{5} x=\cos ^{2} x \\ \sin ^{4} x=1 .\end{array}\right.$
A... | \frac{\pi}{2}(1+4k) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,970 |
$8.4743 \operatorname{tg} 2 x-4 \operatorname{tg} 3 x=\operatorname{tg}^{2} 3 x \cdot \operatorname{tg} 2 x$. | ## Solution.
$$
\begin{aligned}
& 4(\operatorname{tg} 2 x-\operatorname{tg} 3 x)=\operatorname{tg}^{2} 3 x \cdot \operatorname{tg} 2 x+\operatorname{tg} 2 x \Leftrightarrow \frac{-4 \sin x}{\cos 2 x \cdot \cos 3 x}=\frac{\sin 2 x}{\cos 2 x}\left(1+\operatorname{tg}^{2} 3 x\right) \Leftrightarrow \\
& \Leftrightarrow\l... | x_{1}=\pik,x_{2}=\\operatorname{arctg}\sqrt{\frac{3}{5}}+\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,971 |
$8.477 \frac{1+\sin x+\sin ^{2} x+\ldots+\sin ^{n} x+\ldots}{1-\sin x+\sin ^{2} x-\ldots+(-1)^{n} \sin ^{n} x+\ldots}=\frac{4}{1+\operatorname{tg}^{2} x},|\sin x| \neq 1$.
$8.477 \frac{1+\sin x+\sin ^{2} x+\ldots+\sin ^{n} x+\ldots}{1-\sin x+\sin ^{2} x-\ldots+(-1)^{n} \sin ^{n} x+\ldots}=\frac{4}{1+\tan ^{2} x},|\sin... | ## Solution.
$\left\{\begin{array}{l}\frac{1+\sin x}{1-\sin x}=\frac{4}{1+\operatorname{tg}^{2} x}, \\ |\sin x| \neq 1\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{1+\sin x}{1-\sin x}=4 \cos ^{2} x, \\ |\sin x| \neq 1\end{array} \Leftrightarrow\right.\right.$
$$
\Leftrightarrow\left\{\begin{array}{l}
\frac{... | (-1)^{k}\frac{\pi}{6}+\pik | Algebra | proof | Yes | Yes | olympiads | false | 51,972 |
$8.481 \cdot \frac{1}{4} \tan \frac{x}{4}+\frac{1}{2} \tan \frac{x}{2}+\tan x=2 \sqrt{3}+\frac{1}{4} \cot \frac{x}{4}$. | Solution. $\frac{1}{4}\left(\operatorname{tg} \frac{x}{4}-\operatorname{ctg} \frac{x}{4}\right)+\frac{1}{2} \operatorname{tg} \frac{x}{2}+\operatorname{tg} x=2 \sqrt{3} \Leftrightarrow$
$\Leftrightarrow \frac{1}{2} \cdot \frac{\sin ^{2} \frac{x}{4}-\cos ^{2} \frac{x}{4}}{2 \sin \frac{x}{4} \cdot \cos \frac{x}{4}}+\fra... | \frac{\pi}{12}(6k+5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,973 |
$8.482 \frac{3 \tan x-\tan^{3} x}{1-\tan^{2} x} \cdot(\cos 3 x+\cos x)=2 \sin 5 x$. | Solution. $\operatorname{tg} x \cdot \frac{3-\operatorname{tg}^{2} x}{1-\operatorname{tg}^{2} x} \cdot 2 \cos 2 x \cdot \cos x=2 \sin 5 x \Leftrightarrow$
$\Leftrightarrow\left\{\begin{array}{l}\sin x \cdot \cos 2 x \cdot \frac{3 \cos ^{2} x-\sin ^{2} x}{\cos ^{2} x-\sin ^{2} x}=\sin 5 x, \\ \cos x \neq 0\end{array} \... | x_{1}=\pik,x_{2}=\frac{\pi}{8}(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,974 |
$8.483 \operatorname{tg} x-\sin 2 x-\cos 2 x+2\left(2 \cos x-\frac{1}{\cos x}\right)=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$8.483 \tan x - \sin 2x - \cos 2x + 2\left(2 \cos x - \frac{1}{\cos x}\right) = 0$. | Solution. $\frac{\sin x}{\cos x}-2 \sin x \cdot \cos x+2\left(2 \cos x-\frac{1}{\cos x}\right)-\cos 2 x=0 \Leftrightarrow$
$\Leftrightarrow(2-\sin x)\left(2 \cos x-\frac{1}{\cos x}\right)-\cos 2 x=0 \Leftrightarrow$
$\Leftrightarrow(2-\sin x)\left(2 \cos ^{2} x-1\right)-\cos x \cdot \cos 2 x=0 \Leftrightarrow$
$\Lef... | \frac{\pi}{4}(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,975 |
$8.4864 \cos ^{2} 2 x-\operatorname{tg} 4 x=\operatorname{ctg} 2 x$.
$8.4864 \cos ^{2} 2 x-\tan 4 x=\cot 2 x$. | Solution.
$4 \cos ^{2} 2 x=\operatorname{ctg} 2 x+\operatorname{tg} 4 x \Leftrightarrow \frac{\cos 2 x \cdot \cos 4 x+\sin 2 x \cdot \sin 4 x}{\cos 4 x \cdot \sin 2 x}=4 \cos ^{2} 2 x \Leftrightarrow$
$\Leftrightarrow 4 \cos ^{2} 2 x=\frac{\cos 2 x}{\cos 4 x \cdot \sin 2 x} \Leftrightarrow\left\{\begin{array}{l}\cos ... | x_{1}=\frac{\pi}{4}(2k+1),x_{2}=\frac{\pi}{16}(4k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,976 |
$8.487 \frac{\sin ^{2} x-\tan^{2} x}{\cos ^{2} x-\cot^{2} x}-\tan^{6} x+\tan^{4} x-\tan^{2} x=0$. | Solution. $\frac{\sin ^{2} x \cdot\left(1-\frac{1}{\cos ^{2} x}\right)}{\cos ^{2} x \cdot\left(1-\frac{1}{\sin ^{2} x}\right)}-\operatorname{tg}^{6} x+\operatorname{tg}^{4} x-\operatorname{tg}^{2} x=0 \Leftrightarrow$
$\Leftrightarrow\left\{\begin{array}{l}\frac{\sin ^{4} x}{\cos ^{4} x} \cdot \frac{\left(1-\cos ^{2} ... | \frac{\pi}{4}(2k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,977 |
$8.488 \sin ^{10} x+\cos ^{10} x=\frac{29}{64}$. | Solution. Let $\sin ^{2} x=u, \cos ^{2} x=v$.
$\left\{\begin{array}{l}u^{5}+v^{5}=\frac{29}{64}, \Leftrightarrow u^{5}+v^{5}=(u+v)\left(u^{4}-u^{3} v+u^{2} v^{2}-u v^{3}+v^{4}\right)=\left(u^{4}+v^{4}\right)- \\ u+v=1\end{array}\right.$
$-u v \cdot\left(u^{2}+v^{2}\right)+u^{2} v^{2}=\left(u^{2}+v^{2}\right)^{2}-u v \c... | \frac{\pi}{6}(6k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,978 |
$8.490 \tan x + \cot x + \tan^2 x + \cot^2 x + \tan^3 x + \cot^3 x = 6$. | Solution. Let $\operatorname{tg} x+\operatorname{ctg} x=y$. Then
$\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x=y^{2}-2, \operatorname{tg}^{3} x+\operatorname{ctg}^{3} x=(\operatorname{tg} x+\operatorname{ctg} x)\left(\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x-1\right)=y\left(y^{2}-3\right)$.
The original eq... | \frac{\pi}{4}(4k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,979 |
$8.491 \log _{0.5 \sin 2 x} \sin x=\frac{1}{2}$. | ## Solution.
$\log _{\sin x \cdot \cos x} \sin x=\frac{1}{2} \Leftrightarrow \log _{\sin x \cdot \cos x} \sin x=\log _{\sin x \cdot \cos x} \sqrt{\sin x \cdot \cos x} \Leftrightarrow$
$\Leftrightarrow\left\{\begin{array}{l}\sin x>0, \\ \sin x \cdot \cos x \neq 1, \\ \sin x=\sqrt{\sin x \cdot \cos x}\end{array} \Leftr... | \frac{\pi}{4}(8k+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,980 |
$8.492 \log _{\sin x \cdot \cos x} \sin x \cdot \log _{\sin x \cdot \cos x} \cos x=\frac{1}{4}$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,981 | |
$9.216 \log _{5} x+\log _{x} \frac{x}{3}<\frac{\log _{5} x\left(2-\log _{3} x\right)}{\log _{3} x}$. | ## Solution.
$$
\begin{aligned}
& \log _{5} x+1-\log _{x} 30
\end{aligned}
$$
1) Let $x>1$, then $\log _{3} x>0,2 \log _{5} x+1>0$, and the inequality is equivalent to the following:
$\left\{\begin{array}{l}x>1, \\ 1-\log _{3} x>0\end{array} \Leftrightarrow 10$, and the inequality is equivalent to the following:
$\... | x\in(0;\frac{\sqrt{5}}{5})\cup(1;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,983 |
$9.219 \frac{3^{2|x-1|}+3}{4}<3^{|x-1|}$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$9.219 \frac{3^{2|x-1|}+3}{4}<3^{|x-1|}$ | ## Solution.
$3^{2|x-1|}-4 \cdot 3^{|x-1|}+3<0 \Leftrightarrow\left(3^{|x-1|}-1\right)\left(3^{|x-1|}-3\right)<0 \Leftrightarrow$
$\Leftrightarrow 3^{0}<3^{|x-1|}<3^{1} \Leftrightarrow 0<|x-1|<1 \Leftrightarrow\left\{\begin{array}{l}-1<x-1<1, \\ x \neq 1 .\end{array}\right.$
Answer: $x \in(0 ; 1) \cup(1 ; 2)$. | x\in(0;1)\cup(1;2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,985 |
$9.220 \sqrt{x^{2}+3 x+2}-\sqrt{x^{2}-x+1}<1$. | ## Solution.
$$
\begin{aligned}
& \sqrt{x^{2}+3 x+2}<1+\sqrt{x^{2}-x+1} \Leftrightarrow\left\{\begin{array}{l}
x^{2}+3 x+2 \geq 0 \\
x^{2}+3 x+2<x^{2}-x+2+2 \sqrt{x^{2}-x+1}
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
(x+1)(x+2) \geq 0 \\
2 x<\sqrt{x^{2}-x+1}
\end{array}\right.
\end{aligned}
$$
1) ... | (-\infty;-2]\cup[-1;\frac{\sqrt{13}-1}{6}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,986 |
9.221 Prove that among all rectangular parallelepipeds with a given sum of all edges, the one with the greatest volume is a cube. | Solution. Suppose the opposite: there exists a rectangular parallelepiped, not a cube, with a given sum $a=x+y+z$ of its dimensions and with the maximum volume $V_{1}=x y z$. Since this is not a cube, then $\mathrm{x} \neq \mathrm{y}$. Let $\frac{x+y}{2}, \frac{x+y}{2}, z$ be the dimensions of some rectangular parallel... | proof | Geometry | proof | Yes | Yes | olympiads | false | 51,987 |
9.222 For what values of $p$ is the inequality $-9<\frac{3 x^{2}+p x-6}{x^{2}-x+1}<6$ satisfied for all values of $x$? | Solution. The inequality is equivalent to the following inequality:
$-9\left(x^{2}-x+1\right)<0, \\ 3 x^{2}-(p+6) x+12>0 .\end{array}\right.$
Since the solution to this system should be the set $(-\infty ;+\infty)$, each inequality must be satisfied for all $x$, i.e., the discriminants $D_{1}$ and $D_{2}$ are such tha... | p\in(-3;6) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,988 |
### 9.224 Find the domain of the function
$y=\sqrt{\sin x-0.5}+\log _{3}\left(25-x^{2}\right)$. | ## Solution.
$\left\{\begin{array}{l}\sin x \geq \frac{1}{2}, \\ 25-x^{2}>0\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{\pi}{6}+2 \pi n \leq x \leq \frac{5 \pi}{6}+2 \pi n, \\ |x|<5\end{array} \Leftrightarrow\right.\right.$
$\Leftrightarrow\left\{\begin{array}{l}\frac{\pi}{6}+2 \pi n \leq x \leq \frac{5 \pi... | x\in(-5;-\frac{7\pi}{6}]\cup[\frac{\pi}{6};\frac{5\pi}{6}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,990 |
9.225 Prove that for $a>0, b>0, c>0, d>0$ the following is true
$\frac{a+b+c+d}{4} \geq \sqrt[4]{a b c d}$ | Solution. We will use the inequality $A+B \geq 2 \sqrt{A B}(A>0, B>0)$; $(a+b)+(c+d) \geq 2 \sqrt{a b}+2 \sqrt{c d}=2(\sqrt{a b}+\sqrt{c d}) \geq 2 \cdot 2 \sqrt{\sqrt{a b} \cdot \sqrt{c d}}=4 \sqrt[4]{a b c d}$.
The inequality is proved. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 51,991 |
$9.228 \frac{a^{3}+b^{3}}{2} \geq\left(\frac{a+b}{2}\right)^{3},(a>0, b>0)$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$9.228 \frac{a^{3}+b^{3}}{2} \geq\left(\frac{a+b}{2}\right)^{3},(a>0, b>0)$. | Solution.
$$
\begin{aligned}
& \frac{a^{3}+b^{3}}{2}=\frac{a+b}{2}\left(a^{2}-a b+b^{2}\right)=\frac{a+b}{8}\left(4 a^{2}-4 a b+4 b^{2}\right)= \\
& =\frac{a+b}{8}\left(a^{2}+2 a b+b^{2}+3\left(a^{2}-2 a b+b^{2}\right)\right)=\frac{a+b}{8}\left((a+b)^{2}+3(a-b)^{2}\right)= \\
& =\left(\frac{a+b}{2}\right)^{3}+\frac{3(... | Inequalities | proof | Yes | Yes | olympiads | false | 51,993 | |
9.229 \frac{a^{4}+b^{4}}{2} \geq\left(\frac{a+b}{2}\right)^{4},(a>0, b>0)
$$ | Solution. We will prove the auxiliary inequality
$$
\frac{A^{2}+B^{2}}{2} \geq\left(\frac{A+B}{2}\right)^{2}, A, B>0: 2 A^{2}+2 B^{2}=(A+B)^{2}+(A-B)^{2} \geq(A+B)^{2}
$$
- Now we have:
$\frac{a^{4}+b^{4}}{2} \geq\left(\frac{a^{2}+b^{2}}{2}\right)^{2} \geq\left(\left(\frac{a+b}{2}\right)^{2}\right)^{2}=\left(\frac{a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 51,994 |
$9.230 \sqrt[n]{2+\sqrt{3}}+\sqrt[n]{2-\sqrt{3}}>2$. | Solution. Let $a=\sqrt[n]{2+\sqrt{3}}, b=\sqrt[n]{2-\sqrt{3}}$, then
$a+b=(\sqrt{a}-\sqrt{b})^{2}+2 \sqrt{a b}=(\sqrt{a}-\sqrt{b})^{2}+2>2$.
$9.23120$.
2) $3-\left(\log _{3} 2+\log _{2} 3\right)=\left(1-\log _{3} 2\right)+\left(2-\log _{2} 3\right)=$ $=\left(\log _{3} 3-\log _{3} 2\right)+\left(\log _{2} 4-\log _{2} ... | +b>2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,995 |
9.232 Let $x_{1}>0$ be a root of the equation $a x^{2}+b x+c=0$.
Prove: there exists a root $x_{2}$ of the equation $c x^{2}+b x+a=0$ such that $x_{1}+x_{2} \geq 2$. | Solution. By the condition, $a x_{1}^{2}+b x_{1}+c=0 \Rightarrow a+b \cdot \frac{1}{x_{1}}+c \cdot \frac{1}{x_{1}^{2}}=0$,
i.e., $\frac{1}{x_{1}}$ is a root of the equation $c x^{2}+b x+a=0, x_{1}+\frac{1}{x_{1}} \geq 2 \sqrt{x_{1} \cdot \frac{1}{x_{1}}}=2$, which is what we needed to prove. | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,996 |
9.235 The numbers $x_{1}, x_{2}$ are the roots of the equations $5 x^{3}-6=0$ and $6 x^{3}-5=0$ respectively. Prove that $\dot{x}_{1}+x_{2}>2$. | Solution. $x_{1}=\sqrt[3]{\frac{6}{5}}, x_{2}=\sqrt[3]{\frac{5}{6}} \Rightarrow x_{1}+x_{2}=\sqrt[3]{\frac{6}{5}}+\sqrt[3]{\frac{5}{6}}>2 \sqrt{x_{1} x_{2}}=2$, which is what we needed to prove.
## Solve the inequalities $(9.236-9.290)$ : | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,998 |
$9.236 \log _{2}(x-1)-\log _{2}(x+1)+\log _{\frac{x+1}{x-1}} 2>0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$9.236 \log _{2}(x-1)-\log _{2}(x+1)+\log _{\frac{x+1}{x-1}} 2>0$. | Solution. The inequality is equivalent to the following system:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x - 1 > 0 , } \\
{ x + 1 > 0 , } \\
{ \operatorname { log } _ { 2 } \frac { x - 1 } { x + 1 } - \frac { 1 } { \operatorname { log } _ { 2 } \frac { x - 1 } { x + 1 } } > 0 }
\end{array} \Leftrightarrow \l... | x\in(3;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,999 |
$9.237 \log _{x} \log _{2}\left(4^{x}-12\right) \leq 1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$9.237 \log _{x} \log _{2}\left(4^{x}-12\right) \leq 1$. | Solution. $\log _{x} \log _{2}\left(4^{x}-12\right) \leq \log _{x} x$.
1) $\left\{\begin{array}{l}x>1, \\ \log _{2}\left(4^{x}-12\right) \leq x, \\ \log _{2}\left(4^{x}-12\right)>0\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1, \\ 4^{x}-12 \leq 2^{x} \\ 4^{x}-12>1\end{array}, \Leftrightarrow\left\{\begin{array}... | x\in(\log_{4}13;2] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,000 |
$9.243 \sqrt{\log _{1 / 2}\left(x^{2}+4 x-4\right)}<1, x$ - integer. | ## Solution.
The inequality is equivalent to the system:
$\left\{\begin{array}{l}\log _{1 / 2}\left(x^{2}+4 x-4\right) \geq 0, \\ \log _{1 / 2}\left(x^{2}+4 x-4\right)\frac{1}{2}, \\ x-\text { integer }\end{array}, \Leftrightarrow\right.\right.$
^{2}}>1-4 \log _{1 / 8} x$. | Solution. $\sqrt{1-\log _{2}^{2} x}>1+\frac{4}{3} \log _{2} x$.
$$
\text { Let } y=\log _{2} x, \text { then } 3 \sqrt{1-y^{2}}>3+4 y
$$
1) $\left\{\begin{array}{l}3+4 y \geq 0, \\ 9\left(1-y^{2}\right)>(3+4 y)^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}y \geq-\frac{3}{4}, \\ y\left(y+\frac{24}{25}\right)<0... | x\in[\frac{1}{2};1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,002 |
$9.246 \log _{x^{2}}(3-2 x)>1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
$9.246 \log _{x^{2}}(3-2 x)>1$. | Solution. $\log _{x^{2}}(3-2 x)>\log _{x^{2}} x^{2}$.
1) $\left\{\begin{array}{l}x^{2}>1, \\ 3-2 x>x^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}|x|>1, \\ (x-1)(x+3)<0\end{array}, \Leftrightarrow\left\{\begin{array}{c}|x|>1, \\ -3<x<1\end{array} \Leftrightarrow\left\{\begin{array}{c}|x|>1, \\ x \neq 0\end{arr... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,003 |
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