problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
7.290. $\left\{\begin{array}{l}y^{x}=1.5+y^{-x} \\ y^{2.5+x}=64(y>0)\end{array}\right.$ | ## Solution.
Multiply the first equation by $y^{x}$, we have $y^{2 x}-1.5 y^{x}-1=0$. Solving this equation as a quadratic in terms of $y^{x}$, we get $y^{x}=-\frac{1}{2}$ (no solutions), or $y^{x}=2$. From the second equation of the system $y^{2.5} \cdot y^{x}=64 \Rightarrow$ $y^{2.5} \cdot 2=64, y^{2.5}=32, y=4$. Th... | (\frac{1}{2};4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,465 |
7.291. $\left\{\begin{array}{l}\lg (x+y)-\lg 5=\lg x+\lg y-\lg 6, \\ \frac{\lg x}{\lg (y+6)-(\lg y+\lg 6)}=-1 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0, \\ y \neq \frac{6}{5}, \\ y>-6 .\end{array}\right.$
From the condition we have
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ \operatorname { l g } \frac { x + y } { 5 } = \operatorname { l g } \frac { x y } { 6 } , } \\
{ \operatorname { l g }... | (2;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,466 |
7.292. $\left\{\begin{array}{l}\log _{x y} \frac{y}{x}-\log _{y}^{2} x=1, \\ \log _{2}(y-x)=1 .\end{array}\right.$ | ## Solution.
OD3: $\left\{\begin{array}{l}0x .\end{array}\right.$
In the first equation of the system, we transition to the base $y$:
$$
\frac{\log _{y} \frac{y}{x}}{\log _{y} x y}-\log _{y}^{2} x=1 \Leftrightarrow \frac{1-\log _{y} x}{1+\log _{y} x}-\log _{y}^{2} x-1=0 \Leftrightarrow
$$
$\Leftrightarrow \log _{y}... | (1;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,467 |
7.293. $\left\{\begin{array}{l}(x+y)^{x}=(x-y)^{y}, \\ \log _{2} x-\log _{2} y=1\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
(x+y)^{x}=(x-y)^{y}, \\
\log _{2} x-\log _{2} y=1
\end{array}\right.
\] | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0, \\ x \neq \pm y .\end{array}\right.$
From the second equation of the system, we have $\log _{2} \frac{x}{y}=1$, from which $\frac{x}{y}=2$, $x=2 y$. Then from the first equation of the system, we get $(3 y)^{2 y}=y^{y}$, $\left(9 y^{2}\right)^{y}... | (\frac{2}{9};\frac{1}{9}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,468 |
7.294. $\left\{\begin{array}{l}x^{x-2 y}=36, \\ 4(x-2 y)+\log _{6} x=9\end{array}\right.$ (find only integer solutions). | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,469 | |
8.176. $\sin ^{3} x(1+\operatorname{ctg} x)+\cos ^{3} x(1+\operatorname{tg} x)=2 \sqrt{\sin x \cos x}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos x \neq 0, \\ \sin x \cos x>0 .\end{array}\right.$
Write the equation in the form
$$
\begin{aligned}
& \sin ^{3} x \cdot\left(1+\frac{\cos x}{\sin x}\right)+\cos ^{3} x \cdot\left(1+\frac{\sin x}{\cos x}\right)=2 \sqrt{\sin x \cos x} \Left... | \frac{\pi}{4}(8k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,470 |
8.177. $\operatorname{tg}^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2} \operatorname{tg} \frac{x}{2}+\cos ^{2} \frac{x}{2} \operatorname{ctg} \frac{x}{2}+\operatorname{ctg}^{2} \frac{x}{2}+\sin x=4$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos \frac{x}{2} \neq 0, \\ \sin \frac{x}{2} \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
1+\operatorname{tg}^{2} \frac{x}{2}+1+\operatorname{ctg}^{2} \frac{x}{2}+\frac{1-\cos x}{2} \cdot \frac{1-\cos x}{\sin x}+\frac{1+\cos x}{2} \cdot \frac{1+\co... | x_{1}=(-1)^{k+1}\arcsin\frac{2}{3}+\pik;x_{2}=\frac{\pi}{2}(4n+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,471 |
8.178. $\operatorname{tg}\left(120^{\circ}+3 x\right)-\operatorname{tg}\left(140^{\circ}-x\right)=2 \sin \left(80^{\circ}+2 x\right)$. | ## Solution.
Let's write the equation as $\operatorname{tg} 3\left(x+40^{\circ}\right)-\operatorname{tg}\left(180^{\circ}-\left(x+40^{\circ}\right)\right)=$
$=2 \sin 2\left(x+40^{\circ}\right) \Leftrightarrow \operatorname{tg} 3\left(x+40^{\circ}\right)+\operatorname{tg}\left(x+40^{\circ}\right)=2 \sin 2\left(x+40^{\... | -40+60k,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,472 |
8.179. $\sin ^{2} x+2 \sin ^{2} \frac{x}{2}-2 \sin x \sin ^{2} \frac{x}{2}+\operatorname{ctg} x=0$. | ## Solution.
Domain of definition: $\sin x \neq 0$.
$$
\begin{aligned}
& \text { Using the formula } \sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2} \text {, we write } \\
& \sin ^{2} x+1-\cos x-\sin x(1-\cos x)+\frac{\cos x}{\sin x}=0 \Leftrightarrow \\
& \Leftrightarrow \sin ^{2} x+1-\cos x-\sin x+\sin x \cos x+... | \frac{\pi}{4}(4k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,473 |
8.180. $\frac{\cos ^{2} z(1+\operatorname{ctg} z)-3}{\sin z-\cos z}=3 \cos z$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\sin z \neq 0, \\ \sin z-\cos z \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& \cos ^{2} z\left(1+\frac{\cos z}{\sin z}\right)-3=3 \cos z(\sin z-\cos z) \Leftrightarrow \frac{\cos ^{2} z(\sin z+\cos z)}{\sin z}- \\
& -3-3 \cos z(\sin z... | z_{1}=\frac{\pi}{4}(4n-1);z_{2}=\\frac{\pi}{6}+\pik,n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,474 |
8.181. $\frac{1}{2 \operatorname{ctg}^{2} t+1}+\frac{1}{2 \operatorname{tg}^{2} t+1}=\frac{15 \cos 4 t}{8+\sin ^{2} 2 t}$.
8.181. $\frac{1}{2 \cot^{2} t+1}+\frac{1}{2 \tan^{2} t+1}=\frac{15 \cos 4 t}{8+\sin ^{2} 2 t}$. | ## Solution:
Domain of definition: $\left\{\begin{array}{l}\sin t \neq 0, \\ \cos t \neq 0 .\end{array}\right.$
Using the formulas
$$
\operatorname{ctg}^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{1-\cos \alpha}, \operatorname{tg}^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{1+\cos \alpha}, \cos 2 \alpha=2 \cos ^{2} \alp... | \frac{\pi}{12}(6k\1),\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,475 |
8.182. $8 \cos ^{4} x-8 \cos ^{2} x-\cos x+1=0$.
8.182. $8 \cos ^{4} x-8 \cos ^{2} x-\cos x+1=0$.
(No change needed as the text is already in English and is a mathematical equation which is universal.) | ## Solution.
Rewrite the equation as $8\left(\cos ^{2} x\right)^{2}-8 \cos ^{2} x-\cos x+1=0$.
$$
\begin{aligned}
& 8\left(\frac{1}{2}(1+\cos 2 x)\right)^{2}-4(1+\cos 2 x)-\cos x+1=0 \Leftrightarrow 2(1+\cos 2 x)^{2}- \\
& -4(1+\cos 2 x)-\cos x+1=0 \Leftrightarrow \\
& \Leftrightarrow 2+4 \cos 2 x+2 \cos ^{2} 2 x-4-4... | x_{1}=\frac{2}{5}\pin;x_{2}=\frac{2}{3}\pik,wherenk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,476 |
8.183. $\frac{6 \cos ^{3} 2 t+2 \sin ^{3} 2 t}{3 \cos 2 t-\sin 2 t}=\cos 4 t$.
8.183. $\frac{6 \cos ^{3} 2 t+2 \sin ^{3} 2 t}{3 \cos 2 t-\sin 2 t}=\cos 4 t$.
The equation above is already in English and does not require translation. However, if you meant to translate the problem statement into English, it would be:
... | ## Solution.
Domain of definition: $3 \cos 2 t-\sin 2 t \neq 0$.
From the condition we have
$$
\begin{aligned}
& \frac{6 \cos ^{3} 2 t+2 \sin ^{3} 2 t}{3 \cos 2 t-\sin 2 t}-\cos ^{2} 2 t+\sin ^{2} 2 t=0 . \\
& 6 \cos ^{3} 2 t+2 \sin ^{3} 2 t-3 \cos ^{3} 2 t+3 \cos 2 t \sin ^{2} 2 t+\sin 2 t \cos ^{2} 2 t-\sin ^{3} 2... | -\frac{1}{2}\operatorname{arctg}3+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,477 |
8.184. $\cos z \cos 2z \cos 4z \cos 8z=\frac{1}{16}$ | ## Solution.
Multiplying both sides of the equation by $16 \sin z \neq 0$, we get
$8(2 \sin z \cos z) \cos 2 z \cos 4 z \cos 8 z=\sin z \Leftrightarrow 8 \sin 2 z \cos 2 z \cos 4 z \cos 8 z=\sin z$, $4(2 \sin 2 z \cos 2 z) \cos 4 z \cos 8 z=\sin z, 4 \sin 4 z \cos 4 z \cos 8 z=\sin z$, $2(2 \sin 4 z \cos 4 z) \cos 8 ... | z_{1}=\frac{2\pik}{15},k\neq15,z_{2}=\frac{\pi}{17}(2k+1),k\neq17+8,k\inZ,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,478 |
8.185. $\frac{\sin ^{3} \frac{x}{2}-\cos ^{3} \frac{x}{2}}{2+\sin x}=\frac{1}{3} \cos x$. | ## Solution.
From the condition, we get:
$$
\begin{aligned}
& \frac{\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)\left(\sin ^{2} \frac{x}{2}+\sin \frac{x}{2} \cos \frac{x}{2}+\cos ^{2} \frac{x}{2}\right)}{2+2 \sin \frac{x}{2} \cos \frac{x}{2}}-\frac{1}{3}\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)=0 \Le... | \frac{\pi}{2}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,479 |
8.186. $\operatorname{tg}^{2} t-\frac{2 \sin 2 t+\sin 4 t}{2 \sin 2 t-\sin 4 t}=2 \operatorname{ctg} 2 t$.
8.186. $\tan^{2} t-\frac{2 \sin 2 t+\sin 4 t}{2 \sin 2 t-\sin 4 t}=2 \cot 2 t$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \sin t \neq 0, \\ 2 \sin 2 t-\sin 4 t \neq 0 .\end{array}\right.$
From the condition we have:
$$
\begin{aligned}
& \operatorname{tg}^{2} t-\frac{2 \sin 2 t+2 \sin 2 t \cos 2 t}{2 \sin 2 t-2 \sin 2 t \cos 2 t}-2 \operatorname{ctg} 2 t=0 \Lef... | \frac{\pi}{4}(2k+1),k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,480 |
8.187. $\sin ^{2} x \tan x+\cos ^{2} x \cot x+2 \sin x \cos x=\frac{4 \sqrt{3}}{3}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0 \\ \sin x \neq 0\end{array}\right.$
Write the equation in the form:
$$
\begin{aligned}
& \frac{\sin ^{2} x \sin x}{\cos x}+\frac{\cos ^{2} x \cos x}{\sin x}+2 \sin x \cos x-\frac{4 \sqrt{3}}{3}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\sin ^{4... | (-1)^{k}\frac{\pi}{6}+\frac{\pi}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,481 |
8.188. $\operatorname{ctg} x+\operatorname{ctg} 15^{\circ}+\operatorname{ctg}\left(x+25^{\circ}\right)=\operatorname{ctg} 15^{\circ} \operatorname{ctg}\left(x+25^{\circ}\right) \operatorname{ctg} x$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \sin \left(x+25^{\circ}\right) \neq 0 .\end{array}\right.$
## Rewrite the equation as
$$
\frac{\cos x}{\sin x}+\frac{\cos 15^{\circ}}{\sin 15^{\circ}}+\frac{\cos \left(x+25^{\circ}\right)}{\sin \left(x+25^{\circ}\right)}-\frac{\cos 15^{\cir... | 25+90k,k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,482 |
8.189. $\frac{40\left(\sin ^{3} \frac{t}{2}-\cos ^{3} \frac{t}{2}\right)}{16 \sin \frac{t}{2}-25 \cos \frac{t}{2}}=\sin t$. | ## Solution.
Domain of definition: $16 \sin \frac{t}{2}-25 \cos \frac{t}{2} \neq 0$.
## We have
$$
\begin{aligned}
& \frac{40\left(\sin ^{3} \frac{t}{2}-\cos ^{3} \frac{t}{2}\right)}{16 \sin \frac{t}{2}-25 \cos \frac{t}{2}}-2 \sin \frac{t}{2} \cos \frac{t}{2}=0 \Rightarrow 20\left(\sin ^{3} \frac{t}{2}-\cos ^{3} \fr... | 2\operatorname{arctg}\frac{4}{5}+2\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,483 |
8.190. $\frac{(\sin x+\cos x)^{2}-2 \sin ^{2} x}{1+\operatorname{ctg}^{2} x}=\frac{\sqrt{2}}{2}\left(\sin \left(\frac{\pi}{4}-x\right)-\sin \left(\frac{\pi}{4}-3 x\right)\right)$. | ## Solution.
Domain of definition: $\sin x \neq 0$.
Since $\sin \alpha - \sin \beta = 2 \cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}$, we get
$$
\begin{aligned}
& \frac{\sin ^{2} x + 2 \sin x \cos x + \cos ^{2} x - 2 \sin ^{2} x}{1 + \frac{\cos ^{2} x}{\sin ^{2} x}} = \\
& = \frac{\sqrt{2}}{2} \cdot 2... | x_{1}=\frac{\pi}{8}(4k+3);x_{2}=\frac{\pi}{2}(4n+1) | Algebra | proof | Yes | Yes | olympiads | false | 48,484 |
8.191. $\sin ^{-1} t-\sin ^{-1} 2 t=\sin ^{-1} 4 t$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.191. $\sin ^{-1} t-\sin ^{-1} 2 t=\sin ^{-1} 4 t$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\sin t \neq 0, \\ \sin 2 t \neq 0, \\ \sin 4 t \neq 0 .\end{array}\right.$
Rewrite the equation as
$\frac{1}{\sin t}-\frac{1}{\sin 2 t}-\frac{1}{\sin 4 t}=0 \Rightarrow \sin 2 t \sin 4 t-\sin t \sin 4 t-\sin t \sin 2 t=0$.
Using the formula $\sin \alpha \sin \... | \frac{\pi}{7}(2k+1),k\neq7+3,k,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,485 |
8.192. $\frac{1+\sin 2 x}{1-\sin 2 x}+2 \cdot \frac{1+\tan x}{1-\tan x}-3=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin 2 x \neq 1, \\ \operatorname{tg} x \neq 1, \\ \cos x \neq 0 .\end{array}\right.$
From the condition we have:
$$
\begin{aligned}
& \frac{\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x}{\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x}-2 \frac{1+\frac{\sin x}{\cos x}}{1-... | x_{1}=\pik;x_{2}=\operatorname{arctg}2+\pin,wherekn\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,486 |
8.195. $\frac{1-\sin ^{6} z-\cos ^{6} z}{1-\sin ^{4} z-\cos ^{4} z}=2 \cos ^{2} 3 z$. | ## Solution.
Let's write the equation as
$$
\begin{aligned}
& \frac{1-\left(\left(\sin ^{2} z\right)^{3}+\left(\cos ^{2} z\right)^{3}\right)}{1-\left(\sin ^{4} z+\cos ^{4} z\right)}=2 \cos ^{2} 3 z \Leftrightarrow \\
& \Leftrightarrow \frac{1-\left(\sin ^{2} z+\cos ^{2} z\right)\left(\sin ^{4} z-\sin ^{2} z \cos ^{2}... | \frac{\pi}{18}(6k\1),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,487 |
8.196. $\operatorname{ctg} x-\operatorname{tg} x=\frac{\cos x-\sin x}{0.5 \sin 2 x}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos x \neq 0 .\end{array}\right.$
From the condition we have:
$$
\begin{aligned}
& \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}-\frac{\cos x-\sin x}{\sin x \cos x}=0 \Leftrightarrow \frac{\cos ^{2} x-\sin ^{2} x}{\sin x \cos x}-\frac{\cos x... | \frac{\pi}{4}(4k+1),k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,488 |
8.197. $\frac{\cot 2z}{\cot z}+\frac{\cot z}{\cot 2z}+2=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\operatorname{ctg} z \neq 0, \\ \operatorname{ctg} 2 z \neq 0, \\ \sin z \neq 0, \\ \sin 2 z \neq 0 .\end{array}\right.$
From the condition we have:
$$
\begin{aligned}
& \left(\frac{\operatorname{ctg} 2 z}{\operatorname{ctg} z}\right)^{2}+2 \cdot \frac{\oper... | \frac{\pi}{3}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,489 |
8.199. $\frac{\cos x}{\operatorname{ctg}^{2} \frac{x}{2}-\operatorname{tg}^{2} \frac{x}{2}}=\frac{1}{8} \cdot\left(1-\frac{2 \operatorname{ctg} x}{1+\operatorname{ctg}^{2} x}\right)$.
Translate the above text into English, keeping the original text's line breaks and format:
8.199. $\frac{\cos x}{\cot^{2} \frac{x}{2}... | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \sin \frac{x}{2} \neq 0, \\ \cos \frac{x}{2} \neq 0 .\end{array}\right.$
Write the equation in the form:
$\frac{\cos x}{\frac{\cos ^{2} \frac{x}{2}}{\sin ^{2} \frac{x}{2}}-\frac{\sin ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}}}=\frac{1}{8} \cd... | \frac{\pi}{8}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,491 |
8.200. $\frac{3(\cos 2 x+\operatorname{ctg} 2 x)}{\operatorname{ctg} 2 x-\cos 2 x}-2(\sin 2 x+1)=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin 2 x \neq 0, \\ \operatorname{ctg} 2 x-\cos 2 x \neq 0 .\end{array}\right.$
From the condition we have:
$$
\begin{aligned}
& \frac{3\left(\cos 2 x+\frac{\cos 2 x}{\sin 2 x}\right)}{\frac{\cos 2 x}{\sin 2 x}-\cos 2 x}-2(\sin 2 x+1)=0 \Leftrightarrow \\
& ... | (-1)^{k+1}\frac{\pi}{12}+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,492 |
8.201. $\sin 2 x+2 \operatorname{ctg} x=3$.
8.201. $\sin 2 x+2 \cot x=3$. | ## Solution.
Domain of definition: $\sin x \neq 0$.
Since $\sin 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1+\operatorname{tg}^{2} \alpha}$, we have
$\frac{2 \operatorname{tg} x}{1+\operatorname{tg}^{2} x}+\frac{2}{\operatorname{tg} x}-3=0 \Rightarrow 3 \operatorname{tg}^{3} x-4 \operatorname{tg}^{2} x+3 \operatorna... | \frac{\pi}{4}(4n+1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,493 |
8.202. $2 \cos 13 x+3 \cos 3 x+3 \cos 5 x-8 \cos x \cos ^{3} 4 x=0$. | ## Solution.
Since
$$
\begin{aligned}
& \cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha, \cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} \\
& \cos \alpha-\cos \beta=2 \sin \frac{\alpha+\beta}{2} \sin \frac{\beta-\alpha}{2}
\end{aligned}
$$
the equation can be written as
$2 \cos 13 x+... | \frac{\pik}{12},k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,494 |
8.203. $(\sin x+\cos x)^{4}+(\sin x-\cos x)^{4}=3-\sin 4 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.203. $(\sin x+\cos x)^{4}+(\sin x-\cos x)^{4}=3-\sin 4 x$. | Solution.
We have:
$\left(\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x\right)^{2}+\left(\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x\right)^{2}=$
$=3-\sin 4 x \Leftrightarrow(1+\sin 2 x)^{2}+(1-\sin 2 x)^{2}=3-\sin 4 x \Leftrightarrow$
$\Leftrightarrow 1+2 \sin 2 x+\sin ^{2} 2 x+1-2 \sin 2 x+\sin ^{2} 2 x=3-\sin 4 x \Leftrig... | \frac{\pi}{16}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,495 |
8.206. $\operatorname{tg}^{4} x+\operatorname{ctg}^{4} x=\frac{82}{9}(\operatorname{tg} x \operatorname{tg} 2 x+1) \cos 2 x$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0 .\end{array}\right.$
Write the equation in the form:
$$
\begin{aligned}
& \left((\operatorname{tg} x+\operatorname{ctg} x)^{2}-2 \operatorname{tg} x \operatorname{tg} x\right)^{2}-2 \operatorname{tg}^{2} x \operatorname{tg}^{2... | \frac{\pi}{6}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,497 |
8.207. $2 \cos ^{6} 2 t-\cos ^{4} 2 t+1.5 \sin ^{2} 4 t-3 \sin ^{2} 2 t=0$. | Solution.
Since $\cos ^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{2}$ and $\sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2}$, we have
$2\left(\frac{1}{2}(1+\cos 4 t)\right)^{3}-\left(\frac{1}{2}(2+\cos 4 t)\right)^{2}+1.5\left(1-\cos ^{2} 4 t\right)-1.5(1-\cos 4 t)=0 \Leftrightarrow$ $\Leftrightarrow \frac{(1+\cos 4... | \frac{\pi}{8}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,498 |
8.208. $\sin 6 x+2=2 \cos 4 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.208. $\sin 6 x+2=2 \cos 4 x$. | Solution.
Write the equation as $\sin 3(2 x)+2-2 \cos 2(2 x)=0$ and applying the formulas $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$ and $\cos 2 \alpha=1-2 \sin ^{2} \alpha$ we get
$3 \sin 2 x-4 \sin ^{3} 2 x+2-2+4 \sin ^{2} 2 x=0 \Leftrightarrow$
$\Leftrightarrow 4 \sin ^{3} 2 x-4 \sin ^{2} 2 x-3 \sin 2 x=0 \... | x_{1}=\frac{\pik}{2},x_{2}=(-1)^{n+1}\frac{\pi}{12}+\frac{\pin}{2},k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,499 |
8.209. $\sin ^{2} t \tan t+\cos ^{2} t \cot t-2 \sin t \cos t=1+\tan t+\cot t$. . | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \sin t \neq 0 .\end{array}\right.$
Write the equation in the form:
$$
\begin{aligned}
& \frac{\sin ^{2} t \sin t}{\cos t}+\frac{\cos ^{2} t \cos t}{\sin t}-2 \sin t \cos t=1+\frac{\sin t}{\cos t}+\frac{\cos t}{\sin t} \Leftrightarrow \\
& \... | (-1)^{k+1}\frac{\pi}{12}+\frac{\pik}{2},k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,500 |
8.211. $\cos x \cos 2 x \sin 3 x=0.25 \sin 2 x$. | ## Solution.
Let's write the equation as:
$\cos x \cos 2 x \sin 3 x - 0.5 \sin x \cos x = 0 \Leftrightarrow \cos x(\cos 2 x \sin 3 x - 0.5 \sin x) = 0$.
## From this, either
$\cos x = 0, x_{1} = \frac{\pi}{2} + \pi k = \frac{\pi}{2}(2 k + 1), k \in \mathbb{Z}$,
## or
$\cos 2 x \sin 3 x - 0.5 \sin x = 0$, or $2 \c... | x_{1}=\frac{\pi}{2}(2k+1),x_{2}=\frac{\pin}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,502 |
8.212. $\cos 9 x-2 \cos 6 x=2$.
8.212. $\cos 9 x-2 \cos 6 x=2$.
(Note: The equation is already in a universal mathematical format and does not change in translation.) | ## Solution.
Let's write the equation as $\cos 3(3 x)-2 \cos 2(3 x)-2=0$ and, applying the formulas $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$ and $\cos 2 \alpha=2 \cos ^{2} \alpha-1$, we have:
$$
\begin{aligned}
& 4 \cos ^{3} 3 x-3 \cos 3 x-2\left(2 \cos ^{2} 3 x-1\right)-2=0 \Leftrightarrow \\
& \Leftrightarr... | x_{1}=\frac{\pi}{6}(2k+1),x_{2}=\frac{2\pi}{9}(3n\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,503 |
8.213. $2 \sin ^{5} 2 t-\sin ^{3} 2 t-6 \sin ^{2} 2 t+3=0$.
8.213. $2 \sin ^{5} 2 t-\sin ^{3} 2 t-6 \sin ^{2} 2 t+3=0$. | ## Solution.
Let's write the equation as:
$$
\begin{aligned}
& \sin ^{3} 2 t\left(2 \sin ^{2} 2 t-1\right)-3\left(2 \sin ^{2} 2 t-1\right)=0 \Leftrightarrow \\
& \Leftrightarrow\left(2 \sin ^{2} 2 t-1\right)\left(\sin ^{3} 2 t-3\right)=0
\end{aligned}
$$
## From this
1) $2 \sin ^{2} 2 t-1=0, \sin 2 t= \pm \frac{\sq... | \frac{\pi}{8}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,504 |
8.215. $\left(\cos ^{-2} 2 x+\operatorname{tg}^{2} 2 x\right)\left(\sin ^{-2} 2 x+\operatorname{ctg}^{2} 2 x\right)=4 \sin ^{-2} 4 x+5$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 2 x \neq 0, \\ \sin 2 x \neq 0 .\end{array}\right.$
Write the equation in the form:
$$
\left(\frac{1}{\cos ^{2} 2 x}+\frac{\sin ^{2} 2 x}{\cos ^{2} 2 x}\right) \cdot\left(\frac{1}{\sin ^{2} 2 x}+\frac{\cos ^{2} 2 x}{\sin ^{2} 2 x}\right)=\frac{4}{\sin ^... | \frac{\pi}{8}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,506 |
8.216. $\sin 3z + \sin^3 z = \frac{3 \sqrt{3}}{4} \sin 2z$. | ## Solution.
Since $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$ and $\sin 2 \alpha=2 \sin \alpha \cos \alpha$, we have:
$3 \sin z-4 \sin ^{3} z+\sin 3 z-\frac{3 \sqrt{3}}{2} \sin z \cos z=0 \Leftrightarrow$
$\Leftrightarrow 3 \sin z-3 \sin ^{3} z-\frac{3 \sqrt{3}}{2} \sin z \cos z=0 \Leftrightarrow$
$\Leftright... | z_{1}=\pik;z_{2}=\frac{\pi}{2}(2n+1);z_{3}=\\frac{\pi}{6}+2\pi,k,n,\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,507 |
8.217. $\left(\cos 2 x+(\cos x+\sin x)^{2}\right)(\tan x+\cot x)=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0 .\end{array}\right.$
The equation is equivalent to two equations: $\cos 2 x+(\cos x+\sin x)^{2}=0$ or $\operatorname{tg} x+\operatorname{ctg} x=0$. Let's write the first equation as
$$
\begin{aligned}
& \cos ^{2} x-\sin ^{2} x+\c... | \frac{\pi}{4}(4n-1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,508 |
8.219. $3(\operatorname{ctg} t-\operatorname{tg} t)+4 \sin 2 t=0$.
8.219. $3(\cot t-\tan t)+4 \sin 2 t=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin t \neq 0, \\ \cos t \neq 0 .\end{array}\right.$
From the condition we have:
$$
\begin{aligned}
& 3(\operatorname{ctg} t-\operatorname{tg} t)+4 \sin 2 t=0 \Leftrightarrow 3 \cdot\left(\frac{\cos t}{\sin t}-\frac{\sin t}{\cos t}\right)+4 \sin 2 t=0 \Leftr... | \frac{\pi}{3}(3n\1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,510 |
8.220. $\frac{1}{\operatorname{tg}^{2} 2 x+\cos ^{-2} 2 x}+\frac{1}{\operatorname{ctg}^{2} 2 x+\sin ^{-2} 2 x}=\frac{2}{3}$.
8.220. $\frac{1}{\tan^{2} 2 x+\cos ^{-2} 2 x}+\frac{1}{\cot^{2} 2 x+\sin ^{-2} 2 x}=\frac{2}{3}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 2 x \neq 0, \\ \sin 2 x \neq 0 .\end{array}\right.$
Write the equation in the form:
$$
\frac{1}{\frac{\sin ^{2} 2 x}{\cos ^{2} 2 x}+\frac{1}{\cos ^{2} 2 x}}+\frac{1}{\frac{\cos ^{2} 2 x}{\sin ^{2} 2 x}+\frac{1}{\sin ^{2} 2 x}}=\frac{2}{3} \Leftrightarrow
$... | \frac{\pi}{8}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,511 |
8.221. $\operatorname{tg} 3 t+\operatorname{tg} t=2 \sin 4 t$.
8.221. $\tan 3t + \tan t = 2 \sin 4t$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \cos 3 t \neq 0 .\end{array}\right.$
Since $\operatorname{tg} \alpha+\operatorname{tg} \beta=\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}$, we can rewrite the equation as $\frac{\sin (3 t+t)}{\cos 3 t \cos t}-2 \sin 4 t=0 \Leftrightarrow ... | t_{1}=\frac{\pik}{4},k\neq4+2;t_{2}=\\frac{1}{2}\arccos\frac{\sqrt{17}-1}{4}+\pin,k,,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,512 |
8.222. $\sin (3 \pi-x)+\tan(\pi+x)=\frac{\cos ^{-1} x-\cos x}{2 \sin x}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos x \neq 0 .\end{array}\right.$
From the condition we have:
$$
\begin{aligned}
& \sin x+\operatorname{tg} x=\frac{\frac{1}{\cos x}-\cos x}{2 \sin x} \Leftrightarrow \sin x+\frac{\sin x}{\cos x}=\frac{1-\cos ^{2} x}{2 \sin x \cos x} \Leftrig... | \frac{2}{3}\pi(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,513 |
8.223. $\frac{1}{2} \sin 4 x \sin x+\sin 2 x \sin x=2 \cos ^{2} x$. | ## Solution.
Let's write this equation as
$\sin 2 x \cos 2 x \sin x+\sin 2 x \sin x=2 \cos ^{2} x \Leftrightarrow$
$$
\begin{aligned}
& \Leftrightarrow 2 \sin x \cos x\left(2 \cos ^{2} x-1\right) \sin x+2 \sin x \cos x \sin x=2 \cos ^{2} x \Leftrightarrow \\
& \Leftrightarrow 2 \cos x\left(2 \cos ^{2} x-1\right) \si... | \frac{\pi}{2}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,514 |
8.225. $\operatorname{tg} t=\frac{\sin ^{2} t+\sin 2 t-1}{\cos ^{2} t-\sin 2 t+1}$. | ## Solution.
Domain of definition: $\cos t \neq 0$.
Since $\sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2}$ and $\cos ^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{2}$, we have
$$
\begin{aligned}
& \operatorname{tg} t-\frac{\frac{1}{2}(1-\cos 2 t)+\sin 2 t-1}{\frac{1}{2}(1+\cos 2 t)-\sin 2 t+1}=0 \Leftrightarrow \op... | t_{1}=\frac{\pi}{4}(4k+1);t_{2}=\operatorname{arctg}\frac{1-\sqrt{5}}{2}+\pin;t_{3}=\operatorname{arctg}\frac{1+\sqrt{5}}{2}+\pi;k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,515 |
8.226. $\frac{\sin 2 t+2 \cos ^{2} t-1}{\cos t-\cos 3 t+\sin 3 t-\sin t}=\cos t$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\sin t \neq 0, \\ \sin 2 t+\cos 2 t \neq 0 .\end{array}\right.$
Since $\cos \alpha-\cos \beta=2 \sin \frac{\alpha+\beta}{2} \sin \frac{\beta-\alpha}{2}, \cos 2 \alpha=2 \cos ^{2} \alpha-1$
$$
\begin{aligned}
& \sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2... | \frac{\pi}{4}(4k+1),k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,516 |
8.227. $\sin t^{2}-\sin t=0$.
8.227. $\sin t^{2}-\sin t=0$. | ## Solution.
Using the formula $\sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$, we get
$$
2 \sin \frac{t^{2}-t}{2} \cos \frac{t^{2}+t}{2}=0
$$
From this,
1) $\sin \frac{t^{2}-t}{2}=0$
2) $\cos \frac{t^{2}+t}{2}=0$
From the first equation, we have
$$
\frac{t^{2}-t}{2}=\pi k, k \i... | t_{1,2}=\frac{1\\sqrt{1+8\pik}}{2};t_{3,4}=\frac{-1\\sqrt{1+4\pi(1+2n)}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,517 |
8.228. $\sin ^{3} z \sin 3 z+\cos ^{3} z \cos 3 z=\cos ^{3} 4 z$. | ## Solution.
Transform the left side of the equation, applying the formulas $\cos \alpha - \cos \beta = 2 \sin \frac{\alpha + \beta}{2} \sin \frac{\beta - \alpha}{2}, \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$, $\cos 2 \alpha = 2 \cos^2 \alpha - 1, \cos^2 \frac{\alpha}{2} = \frac{1 + \cos... | \frac{\pik}{3},k\in\mathbb{Z} | Algebra | proof | Yes | Yes | olympiads | false | 48,518 |
8.229. $2 \sin ^{4} t(\sin 2 t-3)-2 \sin ^{2} t(\sin 2 t-3)-1=0$. | ## Solution.
Let's write the equation as
$$
\begin{aligned}
& (\sin 2 t-3) \cdot 2 \sin ^{2} t\left(\sin ^{2} t-1\right)-1=0 \\
& -(\sin 2 t-3) \cdot 2 \sin ^{2} t \cos ^{2} t-1=0, (\sin 2 t-3) \cdot 4 \sin ^{2} t \cos ^{2} t+2=0 \\
& (\sin 2 t-3) \sin ^{2} 2 t+2=0, \sin ^{3} 2 t-3 \sin ^{2} 2 t+2=0 \\
& \sin ^{3} 2 ... | t_{1}=\frac{\pi}{4}(4k+1);t_{2}=(-1)^{n}\frac{1}{2}\arcsin(1-\sqrt{3})+\frac{\pin}{2},k,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,519 |
8.230. $\cos x \cos 2 x \cos 4 x \cos 8 x=\frac{1}{8} \cos 15 x$. | ## Solution.
We have
$8(2 \sin x \cos x) \cos 2 x \cos 4 x \cos 8 x=2 \cos 15 x \sin x \Leftrightarrow$
$\Leftrightarrow 4(2 \sin 2 x \cos 2 x) \cos 4 x \cos 8 x=2 \cos 15 x \sin x \Leftrightarrow$
$\Leftrightarrow 2 \sin 8 x \cos 8 x=2 \cos 15 x \sin x, \sin 16 x=2 \cos 15 x \sin x$.
Using the formula $\cos \alpha... | \frac{\pik}{14},k\neq14,k,\in\mathbb{Z} | Algebra | proof | Yes | Yes | olympiads | false | 48,520 |
8.231. $2 \sin ^{4} x+1.25 \sin ^{2} 2 x-\cos ^{4} x=\cos 2 x$. | ## Solution.
We have $8\left(\sin ^{2} x\right)^{2}+5 \sin ^{2} 2 x-4\left(\cos ^{2} x\right)^{2}-4 \cos 2 x=0$. Using the formulas
$$
\begin{aligned}
& \sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2} \text { and } \cos ^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{2}, \text { we find } \\
& 8\left(\frac{1}{2}(1-\cos... | \frac{\pi}{6}(6k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,521 |
8.232. $\sin 2 t \cos 2 t\left(\sin ^{4} 2 t+\cos ^{4} 2 t-1\right)=\frac{1}{2} \sin ^{2} 4 t$. | ## Solution.
We have
$$
(2 \sin 2 t \cos 2 t)\left(\left(\sin ^{2} 2 t+\cos ^{2} 2 t\right)^{2}-2 \sin ^{2} 2 t \cos ^{2} 2 t-1\right)-\sin ^{2} 4 t=0 \Leftrightarrow
$$
$\Leftrightarrow \sin 4 t\left(1-2 \sin ^{2} 2 t \cos ^{2} 2 t-1\right)-\sin ^{2} 4 t=0 \Leftrightarrow$
$\Leftrightarrow-\sin 4 t \cdot 4 \sin ^{... | \frac{\pik}{4},k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,522 |
8.233. $\sin 2 x-2 \cos ^{2} x+4(\sin x-\cos x+\operatorname{tg} x-1)=0$.
8.233. $\sin 2 x-2 \cos ^{2} x+4(\sin x-\cos x+\tan x-1)=0$. | ## Solution.
Domain of definition: $\cos x \neq 0$.
Let's write the equation in the form
$$
\begin{aligned}
& \left(2 \sin x \cos x-2 \cos ^{2} x\right)+4\left((\sin x-\cos x)+\frac{\sin x}{\cos x}-1\right)=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos x(\sin x-\cos x)+4\left((\sin x-\cos x)+\frac{\sin x-\cos x}{\co... | \frac{\pi}{4}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,523 |
8.234. $\frac{1}{2}\left(\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x\right)=1+\frac{2}{\sqrt{3}} \operatorname{ctg} 2 x$.
8.234. $\frac{1}{2}\left(\tan^{2} x+\cot^{2} x\right)=1+\frac{2}{\sqrt{3}} \cot 2 x$. | ## Solution.
Domain of definition: $\sin 2 x \neq 0$.
## Write the equation in the form
$1+1+\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x=4+\frac{4 \operatorname{ctg} 2 x}{\sqrt{3}} \Leftrightarrow$
$\Leftrightarrow \frac{1}{\cos ^{2} x}+\frac{1}{\sin ^{2} x}=\frac{4(\sqrt{3}+\operatorname{ctg} 2 x)}{\sqrt{3}} ... | x_{1}=\frac{\pi}{4}(2k+1);x_{2}=\frac{\pi}{6}(3n+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,524 |
8.235. $\operatorname{ctg}^{4} x=\cos ^{3} 2 x+1$.
8.235. $\cot^{4} x=\cos ^{3} 2 x+1$. | ## Solution.
Domain of definition: $\sin x \neq 0$.
Since $1+\operatorname{ctg}^{2} \alpha=\frac{1}{\sin ^{2} a}$, the equation becomes $\left(\frac{1}{\sin ^{2} x}-1\right)^{2}=\cos ^{3} 2 x+1 \Leftrightarrow \frac{1}{\sin ^{4} x}-\frac{2}{\sin ^{2} x}+1=\cos ^{3} 2 x+1 \Leftrightarrow$ $\Leftrightarrow \frac{1-2 \s... | x_{1}=\frac{\pi}{4}(2k+1);x_{2}=\frac{\pi}{2}(2n+1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,525 |
8.236. $\frac{1}{\sin ^{3} \frac{x}{2} \cos ^{3} \frac{x}{2}}-6 \cos ^{-1} x=\tan^{3} \frac{x}{2}+\cot^{3} \frac{x}{2}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos x \neq 0 .\end{array}\right.$
Since $\sin \alpha=2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}, \operatorname{tg} \frac{\alpha}{2}=\frac{1-\cos \alpha}{\sin \alpha}, \operatorname{ctg} \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}$... | \frac{\pi}{4}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,526 |
8.237. $4 \sin 2 x \sin 5 x \sin 7 x - \sin 4 x = 0$. | ## Solution.
Rewrite the equation as
$4 \sin 2 x \sin 5 x \sin 7 x - 2 \sin 2 x \cos 2 x = 0$,
$$
\begin{aligned}
& 2 \sin 2 x (2 \sin 5 x \sin 7 x - \cos 2 x) = 0 \Leftrightarrow \\
& \sin 2 x (\cos 2 x - \cos 12 x - \cos 2 x) = 0, \sin 2 x \cos 12 x = 0
\end{aligned}
$$
From this,
1) $\sin 2 x = 0, 2 x = \pi k, ... | x_{1}=\frac{\pik}{2};x_{2}=\frac{\pi}{24}(2n+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,527 |
8.238. $\sin x+\cos x+\sin 2 x+\sqrt{2} \sin 5 x=\frac{2 \cot x}{1+\cot^{2} x}$. | ## Solution.
Since $\sin \alpha+\cos \alpha=\sqrt{2} \cos \left(\frac{\pi}{4}-\alpha\right)$ and $\frac{2 \operatorname{ctg} \alpha}{1+\operatorname{ctg}^{2} \alpha}=\sin 2 \alpha$, we have $\sqrt{2} \cos \left(\frac{\pi}{4}-x\right)+\sqrt{2} \sin 5 x+\sin 2 x=\sin 2 x \Leftrightarrow$ $\Leftrightarrow \sqrt{2} \cos \... | x_{1}=-\frac{\pi}{24}+\frac{\pik}{3};x_{2}=\frac{5\pi}{16}+\frac{\pin}{2},k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,528 |
8.239. $3 \sin ^{2} \frac{x}{2} \cos \left(\frac{3 \pi}{2}+\frac{x}{2}\right)+3 \sin ^{2} \frac{x}{2} \cos \frac{x}{2}-\sin \frac{x}{2} \cos ^{2} \frac{x}{2}=\sin ^{2}\left(\frac{\pi}{2}+\frac{x}{2}\right) \cos \frac{x}{2}$. | ## Solution.
From the condition we have
$$
3 \sin ^{2} \frac{x}{2} \sin \frac{x}{2} + 3 \sin ^{2} \frac{x}{2} \cos \frac{x}{2} - \sin \frac{x}{2} \cos ^{2} \frac{x}{2} - \cos ^{2} \frac{x}{2} \cos \frac{x}{2} = 0
$$
$3 \sin ^{2} \frac{x}{2} \left( \sin \frac{x}{2} + \cos \frac{x}{2} \right) - \cos ^{2} \frac{x}{2} \... | x_{1}=\frac{\pi}{2}(4k-1);x_{2}=\frac{\pi}{3}(6\1) | Algebra | proof | Yes | Yes | olympiads | false | 48,529 |
8.240. $\operatorname{tg}\left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot \frac{1+\sin x}{\sin x}=\sqrt{2} \cos x$.
8.240. $\tan\left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot \frac{1+\sin x}{\sin x}=\sqrt{2} \cos x$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos \left(\frac{\pi}{4}-\frac{x}{2}\right) \neq 0 .\end{array}\right.$
Since $\operatorname{tg} \frac{\alpha}{2}=\frac{1-\cos \alpha}{\sin \alpha}$, we have
$\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)} \cd... | x_{1}=\frac{\pi}{2}(4k+1);x_{2}=(-1)^{n}\frac{\pi}{4}+\pin,k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,530 |
8.241. $\operatorname{tg}^{3} z+\operatorname{ctg}^{3} z-8 \sin ^{-3} 2 z=12$.
8.241. $\tan^{3} z+\cot^{3} z-8 \sin^{-3} 2 z=12$. | ## Solution.
Domain of definition: $\sin 2 z \neq 0$.
Rewrite the equation as
$$
\operatorname{tg}^{3} z+\operatorname{ctg}^{3} z-\frac{8}{\left(\frac{2 \operatorname{ctg} z}{1+\operatorname{ctg}^{2} z}\right)^{3}}=12 \Leftrightarrow
$$
$$
\begin{aligned}
& \Leftrightarrow \operatorname{tg}^{3} z+\operatorname{ctg}... | (-1)^{k+1}\frac{\pi}{12}+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,531 |
8.242. $\frac{1}{\tan 5 x+\tan 2 x}-\frac{1}{\cot 5 x+\cot 2 x}=\tan 3 x$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 3 x \neq 0, \\ \operatorname{tg} 5 x+\operatorname{tg} 2 x \neq 0, \\ \operatorname{ctg} 5 x+\operatorname{ctg} 2 x \neq 0, \\ \cos 5 x \neq 0, \\ \cos 2 x \neq 0, \\ \sin 5 x \neq 0, \\ \sin 2 x \neq 0 .\end{array}\right.$
Since $\operatorname{tg} \alpha+\... | \frac{\pi}{20}(2k+1),k\neq5+2,k | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,532 |
8.244. $\operatorname{ctg}^{4} x=\cos ^{2} 2 x-1$.
8.244. $\cot^{4} x=\cos ^{2} 2 x-1$. | ## Solution.
Domain of definition: $\sin x \neq 0$.
From the condition we have
$$
\begin{aligned}
& \frac{\left(\cos ^{2} x\right)^{2}}{\left(\sin ^{2} x\right)^{2}}+1-\cos ^{2} 2 x=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\left(\frac{1}{2}(1+\cos 2 x)\right)^{2}}{\left(\frac{1}{2}(1-\cos 2 x)\right)^{2}}+1-\cos... | \frac{\pi}{2}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,534 |
8.245. $\frac{4 \sin ^{2} \frac{t}{2}-1}{\cos t}=\tan t(1-2 \cos t)$. | Solution.
Domain of definition: $\cos t \neq 0$.
Since $\sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2}$, we have:
$$
\begin{aligned}
& \frac{2(1-\cos t)-1}{\cos t}-\frac{\sin t(1-2 \cos t)}{\cos t}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{1-2 \cos t-\sin t(1-2 \cos t)}{\cos t}=0 \Leftrightarrow(1-2 \cos t)(1... | \frac{\pi}{3}(6k\1),k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,535 |
8.246. $3 \sin ^{2} z \cos ^{2}\left(\frac{\pi}{2}+z\right)-\frac{1}{2} \sin ^{2} 2 z-5 \cos ^{4} z+2 \cos 2 z=0$. | ## Solution.
From the condition we have:
$6\left(\sin ^{2} z\right)^{2}-\left(1-\cos ^{2} 2 z\right)-10\left(\cos ^{2} z\right)^{2}+4 \cos 2 z=0$.
Since $\sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2}$ and $\cos ^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{2}$, we get
$6 \cdot \frac{1}{4}(1-\cos 2 z)^{2}-1+\cos ^... | \frac{\pi}{3}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,536 |
8.247. $\frac{\cos ^{3} 3 t}{\tan t}+\frac{\cos ^{2} t}{\tan 3 t}=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \cos 3 t \neq 0, \\ \tan t \neq 0, \\ \tan 3 t \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& \frac{\cos ^{2} 3 t \cos t}{\sin t}+\frac{\cos ^{2} t \cos 3 t}{\sin 3 t}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\cos... | \frac{\pi}{4}(2+1),\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,537 |
8.249. $\frac{\cos ^{4} 2 x+\sin ^{4} 2 x}{\cos ^{4} 2 x-\sin ^{4} 2 x}-\frac{1}{2} \cos 4 x=\frac{\sqrt{3}}{2} \sin ^{-1} 4 x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.249. $\frac{\cos ^{4} 2 x+\sin ^{4} 2 x}{\cos ... | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin 4 x \neq 0, \\ \cos ^{4} 2 x-\sin ^{4} 2 x \neq 0 .\end{array}\right.$
From the condition we have
$$
\begin{aligned}
& \frac{\left(\cos ^{2} 2 x+\sin ^{2} 2 x\right)^{2}-2 \cos ^{2} 2 x \cos ^{2} 2 x}{\left(\cos ^{2} 2 x+\sin ^{2} 2 x\right)\left(\cos ^... | \frac{\pi}{12}(3k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,539 |
8.252. $(\sin x-\cos x)^{2}+\tan x=2 \sin ^{2} x$. | ## Solution.
Domain of definition: $\cos x \neq 0$.
Since $\operatorname{tg} \alpha=\frac{1-\cos 2 \alpha}{\sin 2 \alpha}$ and $\sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2}$, we have $\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x+\frac{1-\cos 2 x}{\sin 2 x}-(1-\cos 2 x)=0 \Leftrightarrow$ $\Leftrightarrow 1-\sin 2 x+... | \frac{\pi}{4}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,542 |
8.253. $\sin 3 t-\sin t=\frac{8 \cos t \cot 2 t}{4-\sin^{-2} t}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin 2 t \neq 0, \\ \sin t \neq \pm \frac{1}{2}\end{array}\right.$.
Since $\sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$, we have
$$
2 \sin t \cos 2 t=\frac{\frac{8 \cos t \cos 2 t}{\sin 2 t}}{4-\frac{1}{\sin ^{2} t}} \Lef... | t_{1}=\frac{\pi}{4}(2k+1),t_{2}=\\frac{\pi}{3}+\pi,k,\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,543 |
8.254. $\sin ^{2} 2 x \cos \left(\frac{3 \pi}{2}-2 x\right)+3 \sin 2 x \sin ^{2}\left(\frac{3 \pi}{2}+2 x\right)+2 \cos ^{3} 2 x=0$. | ## Solution.
From the condition we have
$$
\begin{aligned}
& -\sin ^{2} 2 x \sin 2 x+3 \sin 2 x \cos ^{2} 2 x+2 \cos ^{3} 2 x=0 \Leftrightarrow \\
& \Leftrightarrow \sin ^{3} 2 x-3 \sin 2 x \cos ^{2} 2 x-2 \cos ^{3} 2 x=0 \Leftrightarrow \\
& \Leftrightarrow \operatorname{tg}^{3} 2 x-3 \operatorname{tg} 2 x-2=0 \Left... | x_{1}=\frac{\pi}{8}(4k-1);x_{2}=\frac{1}{2}\operatorname{arctg}2+\frac{\pin}{2},k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,544 |
8.255. $\operatorname{tg}(x+1) \operatorname{ctg}(2 x+3)=1$.
8.255. $\tan(x+1) \cot(2 x+3)=1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos (x+1) \neq 0, \\ \sin (2 x+3) \neq 0 .\end{array}\right.$
From the condition we have
$$
\begin{aligned}
& \operatorname{tg}(x+1)=\operatorname{tg}(2 x+3) \Leftrightarrow \operatorname{tg}(x+1)-\operatorname{tg}(2 x+3)=0 \Leftrightarrow \\
& \Leftrightar... | -2+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,545 |
8.257. $\operatorname{tg}^{2} \frac{z}{2}+\operatorname{ctg}^{2} \frac{z}{2}-2=4 \operatorname{tg} z$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos z \neq 0, \\ \sin z \neq 0 .\end{array}\right.$
Write the equation in the form
$$
\begin{aligned}
& \left(\operatorname{tg} \frac{z}{2}+\operatorname{ctg} \frac{z}{2}\right)^{2}-2 \operatorname{tg} \frac{z}{2} \operatorname{ctg} \frac{z}{2}-2-4 \operato... | \frac{\pi}{4}(4n+1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,547 |
8.258. $\cos ^{3} z \cos 3 z+\sin ^{3} z \sin 3 z=\frac{\sqrt{2}}{4}$. | ## Solution.
Let's write the equation as
$\cos ^{2} z(2 \cos z \cos 3 z)+\sin ^{2} z(2 \sin z \sin 3 z)=\frac{\sqrt{2}}{2}$.
Since
$$
\cos \alpha \cos \beta=\frac{1}{2}(\cos (\alpha-\beta)+\cos (\alpha+\beta)) \text { and }
$$
$\sin \alpha \sin \beta=\frac{1}{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta))$, we have
... | \frac{\pi}{8}(8k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,548 |
8.259. $\operatorname{ctg} x=\frac{\sin ^{2} x-2 \sin ^{2}\left(\frac{\pi}{4}-x\right)}{\cos ^{2} x+2 \cos ^{2}\left(\frac{\pi}{4}+x\right)}$.
8.259. $\operatorname{cot} x=\frac{\sin ^{2} x-2 \sin ^{2}\left(\frac{\pi}{4}-x\right)}{\cos ^{2} x+2 \cos ^{2}\left(\frac{\pi}{4}+x\right)}$. | Solution.
Domain of definition: $\sin x \neq 0,\left\{\begin{array}{l}\cos x \neq 0, \\ \cos \left(\frac{\pi}{4}+x\right) \neq 0\end{array} \Leftrightarrow x \neq \pi n,\left\{\begin{array}{l}x \neq \frac{\pi}{2}+\pi n, \\ x \neq \frac{\pi}{4}+\pi n\end{array} \Leftrightarrow x \neq \pi n\right.\right.$.
Rewrite the ... | x_{1}=\frac{\pi}{4}(4+1),x_{2}=-\operatorname{arctg}2+\pin,x_{3}=\frac{\pi}{2}(2+1),\text{where},n\text{}\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,549 |
8.261. $(2 \cos 2 t+5) \cos ^{4} t-(2 \cos 2 t+5) \sin ^{4} t=3$. | ## Solution.
From the condition we have:
$$
\begin{aligned}
& (2 \cos 2 t+5)\left(\cos ^{4} t-\sin ^{4} t\right)-3=0 \Leftrightarrow \\
& \Leftrightarrow(2 \cos 2 t+5)\left(\cos ^{2} t+\sin ^{2} t\right)\left(\cos ^{2} t-\sin ^{2} t\right)-3=0 \Leftrightarrow \\
& \Leftrightarrow(2 \cos 2 t+5) \cos 2 t-3=0 \Leftright... | \frac{\pi}{6}(6k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,550 |
8.262. $\operatorname{tg} z \operatorname{tg}\left(z+60^{\circ}\right) \operatorname{tg}\left(z+120^{\circ}\right)=\sqrt{3}$. | Solution.
$$
\text { Domain of definition: }\left\{\begin{array}{l}
\cos z \neq 0 \\
\cos \left(z+60^{\circ}\right) \neq 0 \\
\cos \left(z+120^{\circ}\right) \neq 0
\end{array}\right.
$$
Write the equation in the form
$\frac{\sin z \sin \left(z+60^{\circ}\right) \sin \left(z+120^{\circ}\right)}{\cos z \cos \left(z+6... | -20+60k,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,551 |
8.263. $\cos 3 x+\cos \frac{5 x}{2}=2$.
8.263. $\cos 3 x+\cos \frac{5 x}{2}=2$. | ## Solution.
The equation is equivalent to a system of two equations
$$
\left\{\begin{array} { l }
{ \operatorname { cos } 3 x \neq 1 , } \\
{ \operatorname { cos } \frac { 5 x } { 2 } = 1 , }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ 3 x = 2 \pi k , k \in Z , } \\
{ \frac { 5 x } { 2 } = 2 \pi n , n... | 4\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,552 |
8.264. $1-\frac{2(\cos 2t-\tan t \sin 2t)}{\cos^{-2} t}=\sin^4 t-\cos^4 t$. | ## Solution.
Domain of definition: $\cos t \neq 0$.
## Rewrite the equation as
$1-\frac{2\left(\cos 2 t-\frac{\sin t}{\cos t} \cdot 2 \sin t \cos t\right)}{\frac{1}{\cos ^{2} t}}+\left(\sin ^{4} t-\cos ^{4} t\right)=0 \Leftrightarrow$
$\Leftrightarrow 1-2\left(\cos 2 t-2 \sin ^{2} t\right) \cos ^{2} t+\left(\cos ^{... | \pik,k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,553 |
8.265. $2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)=\cos 2 x$. | ## Solution.
We have
$2\left(\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}\right)-3\left(\left(\sin ^{2} x\right)^{2}+\left(\cos ^{2} x\right)^{2}\right)-\cos 2 x=0 \Leftrightarrow$
$\Leftrightarrow 2\left(\sin ^{2} x+\cos ^{2} x\right) \cdot\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)-3\... | \frac{\pi}{2}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,554 |
8.266. $\cos ^{3} x+\frac{1}{2} \sin 2 x-\cos x \sin ^{3} x+4 \sin x+4=0$. | ## Solution.
Let's rewrite the equation as
$\cos ^{3} x+\sin x \cos x-\cos x \sin x \sin ^{2} x+4(\sin x+1)=0 \Leftrightarrow$
$\Leftrightarrow \cos ^{3} x+\sin x \cos x\left(1-\sin ^{2} x\right)+4(\sin x+1)=0 \Leftrightarrow$
$\Leftrightarrow \cos ^{3} x+\sin x \cos ^{3} x+4(\sin x+1)=0 \Leftrightarrow$
$\Leftrig... | \frac{\pi}{2}(4k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,555 |
8.267. $\frac{2\left(\cos ^{3} x+2 \sin ^{3} x\right)}{2 \sin x+3 \cos x}=\sin 2 x$. | ## Solution.
Domain of definition: $2 \sin x + 3 \cos x \neq 0$.
## Write the equation as
$$
\begin{aligned}
& \frac{2\left(\cos ^{3} x + 2 \sin ^{3} x\right)}{2 \sin x + 3 \cos x} - 2 \sin x \cos x = 0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{3} x + 4 \sin ^{3} x - 4 \sin ^{2} x \cos x - 6 \sin x \cos ^{2} x =... | x_{1}=\frac{\pi}{4}(4k-1);x_{2}=\operatorname{arctg}(1-\frac{\sqrt{2}}{2})+\pin;x_{3}=\operatorname{arctg}(1+\frac{\sqrt{2}}{2})+\pi, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,556 |
8.268. $\operatorname{tg} \frac{3 x}{2}-\operatorname{tg} \frac{x}{2}=2 \sin x$.
8.268. $\tan \frac{3 x}{2}-\tan \frac{x}{2}=2 \sin x$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos \frac{3 x}{2} \neq 0, \\ \cos \frac{x}{2} \neq 0 .\end{array}\right.$
By the formula $\operatorname{tg} \alpha-\operatorname{tg} \beta=\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta}$, we have
$\frac{\sin x}{\cos \frac{3 x}{2} \cos \frac{x}{2}}-2 \si... | x_{1}=2\pik,x_{2}=\\arccos\frac{-1+\sqrt{17}}{4}+2\pin,k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,557 |
8.269. $\frac{1+\sin 2 x}{\cos 2 x}+\frac{1+\tan x \tan \frac{x}{2}}{\cot x+\tan \frac{x}{2}}=1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 2 x \neq 0, \\ \cos x \neq 0, \\ \cos \frac{x}{2} \neq 0, \\ \sin x \neq 0,\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& \frac{\cos ^{2} x+2 \sin x \cos x+\sin ^{2} x}{\cos ^{2} x-\sin ^{2} x}+\frac{1+\frac{\sin x \sin \frac{x}{2}}{\c... | \operatorname{arctg}3+\pik,k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,558 |
8.270. $\sin ^{3} x \cos 3 x+\cos ^{3} x \sin 3 x+0.375=0$. | ## Solution.
We have $\sin ^{2} x(2 \sin x \cos 3 x)+\cos ^{2} x(2 \cos x \sin 3 x)+2 \cdot 0.375=0$. Using the formula $\sin \alpha \cos \beta=\frac{1}{2}(\sin (\alpha-\beta)+\sin (\alpha+\beta))$, we find
$$
\begin{aligned}
& \sin ^{2} x(-\sin 2 x+\sin 4 x)+\cos ^{2} x(\sin 2 x+\sin 4 x)+0.75=0 \Leftrightarrow \\
&... | (-1)^{k+1}\frac{\pi}{24}+\frac{\pik}{4},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,559 |
8.272. $\sin ^{3} 2 t+\cos ^{3} 2 t+\frac{1}{2} \sin 4 t=1$. | ## Solution.
Let's write the equation in the form
$$
(\sin 2 t+\cos 2 t)\left(\sin ^{2} 2 t-\sin 2 t \cos 2 t+\cos ^{2} 2 t\right)+\sin 2 t \cos 2 t-1=0 \Leftrightarrow
$$
$\Leftrightarrow(\sin 2 t+\cos 2 t)(1-\sin 2 t \cos 2 t)-(1-\sin 2 t \cos 2 t)=0 \Leftrightarrow$
$\Leftrightarrow(1-\sin 2 t \cos 2 t)(\sin 2 t... | t_{1}=\pik,t_{2}=\frac{\pi}{4}(4n+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,561 |
8.273. $\operatorname{tg} z \operatorname{tg} 2 z=\operatorname{tg} z+\operatorname{tg} 2 z$.
8.273. $\tan z \tan 2z = \tan z + \tan 2z$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\cos z \neq 0 \\ \cos 2 z \neq 0 .\end{array}\right.$
Rewrite the equation as
$\frac{\sin z \sin 2 z}{\cos z \cos 2 z}=\frac{\sin z}{\cos z}+\frac{\sin 2 z}{\cos 2 z} \Leftrightarrow$
$\Leftrightarrow \frac{\sin z \sin 2 z}{\cos z \cos 2 z}=\frac{\sin z \co... | z_{1}=\pik,z_{2}=-\operatorname{arctg}3+\pin,k,n\inZ | Algebra | proof | Yes | Yes | olympiads | false | 48,562 |
8.274. $\frac{\sin ^{3} x+\cos ^{3} x}{2 \cos x-\sin x}=\cos 2 x$.
8.274. $\frac{\sin ^{3} x+\cos ^{3} x}{2 \cos x-\sin x}=\cos 2 x$.
(No change in the formula as it is already in a universal mathematical notation that does not require translation.) | ## Solution.
Domain of definition: $2 \cos x - \sin x \neq 0$.
From the condition, we have
$$
\begin{aligned}
& \frac{\sin ^{3} x + \cos ^{3} x}{2 \cos x - \sin x} - \cos ^{2} x + \sin ^{2} x = 0 \Leftrightarrow \\
& \Leftrightarrow \sin ^{3} x + \cos ^{3} x - 2 \cos ^{3} x + 2 \cos x \sin ^{2} x + \sin x \cos ^{2} ... | x_{1}=\frac{\pi}{2}(2k+1),x_{2}=\frac{\pi}{4}(4n-1),x_{3}=\operatorname{arctg}\frac{1}{2}+\pi,k,n,\in\mathbb{Z} | Algebra | proof | Yes | Yes | olympiads | false | 48,563 |
8.275. $\frac{\cot 4 t}{\sin ^{2} t}+\frac{\cot t}{\sin ^{2} 4 t}=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\sin t \neq 0, \\ \sin 4 t \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
\frac{\frac{\cos 4 t}{\sin 4 t}}{\sin ^{2} t}+\frac{\frac{\cos t}{\sin t}}{\sin ^{2} 4 t}=0, \frac{\cos 4 t}{\sin 4 t \sin ^{2} t}+\frac{\cos t}{\sin t \sin ^{2} 4 t}=0 \Leftright... | t_{1}=\frac{\pik}{5},k\neq5;t_{2}=\frac{\pi}{6}(2n+1),n\neq3+1,k,n,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,564 |
8.276. $\operatorname{tg}^{4} x=36 \cos ^{2} 2 x$.
8.276. $\tan^{4} x=36 \cos ^{2} 2 x$. | ## Solution.
Domain of definition: $\cos x \neq 0$.
From the condition, we have
$$
\begin{aligned}
& \frac{\left(\sin ^{2} x\right)^{2}}{\left(\cos ^{2} x\right)^{2}}-36 \cos ^{2} 2 x=0 \Leftrightarrow \frac{(1-\cos 2 x)^{2}}{(1+\cos 2 x)^{2}}-36 \cos ^{2} 2 x=0 \Leftrightarrow \\
& \Leftrightarrow(1-\cos 2 x)^{2}-3... | x_{1}=\\frac{\pi}{3}+\pik,x_{2}=\\frac{1}{2}\arccos(-\frac{1}{3})+\pin,x_{3}=\\frac{1}{2}\arccos\frac{-7+\sqrt{73}}{12}+\pi,\text{where}k,n\text{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,565 |
8.277. $\operatorname{ctg} x-\operatorname{tg} x-2 \operatorname{tg} 2 x-4 \operatorname{tg} 4 x+8=0$.
8.277. $\cot x - \tan x - 2 \tan 2x - 4 \tan 4x + 8 = 0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos x \neq 0, \\ \cos 2 x \neq 0, \\ \cos 4 x \neq 0 .\end{array}\right.$
Rewrite the equation as
$$
\begin{aligned}
& \left(\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}\right)-2 \tan 2 x-4 \tan 4 x+8=0 \Leftrightarrow \\
& \Leftrightarrow ... | \frac{\pi}{32}(4k+3),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,566 |
8.278. $4 \sin ^{3} x \cos 3 x+4 \cos ^{3} x \sin 3 x=3 \sin 2 x$.
8.278. $4 \sin ^{3} x \cos 3 x + 4 \cos ^{3} x \sin 3 x = 3 \sin 2 x$. | ## Solution.
Let's rewrite the equation as
$2 \sin ^{2} x(2 \sin x \cos 3 x)+2 \cos ^{2} x(2 \sin 3 x \cos x)-3 \sin 2 x=0 \Leftrightarrow$
$\Leftrightarrow 2 \sin ^{2} x(-\sin 2 x+\sin 4 x)+2 \cos ^{2} x(\sin 2 x+\sin 4 x)-3 \sin 2 x=0 \Leftrightarrow$ $\Leftrightarrow-2 \sin ^{2} x \sin 2 x+2 \sin ^{2} x \sin 4 x+... | x_{1}=\frac{\pi}{6}(2n+1),n\inZ;x_{2}=\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,567 |
8.279. $2 \cos z \sin ^{3}\left(\frac{3 \pi}{2}-z\right)-5 \sin ^{2} z \cos ^{2} z+\sin z \cos ^{3}\left(\frac{3 \pi}{2}+z\right)=\cos 2 z$. | ## Solution.
From the condition we have
$-2 \cos z \cos ^{3} z-5 \sin ^{2} z \cos ^{2} z+\sin z \sin ^{3} z-\cos 2 z=0 \Leftrightarrow$
$\Leftrightarrow-2 \cos ^{4} z-5 \sin ^{2} z \cos ^{2} z+\sin ^{4} z-\cos 2 z=0 \Leftrightarrow$
$\Leftrightarrow-2\left(1-\sin ^{2} z\right)^{2}-5 \sin ^{2} z\left(1-\sin ^{2} z\r... | \frac{\pi}{3}(3n\1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,568 |
8.280. $\sin 2 x \sin 6 x \cos 4 x+\frac{1}{4} \cos 12 x=0$. | ## Solution.
From the condition we have
$$
\begin{aligned}
& (2 \sin 2 x \sin 6 x) \cos 4 x+\cos 12 x=0 \Leftrightarrow \\
& \Leftrightarrow 2(\cos 4 x-\cos 8 x) \cos 4 x+\cos 12 x=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{2} 4 x-2 \cos 8 x \cos 4 x+\cos 12 x=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{2} 4... | x_{1}=\frac{\pi}{8}(2k+1);x_{2}=\frac{\pi}{12}(6n\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,569 |
8.281. $2 \sin 2 x+3 \tan x=5$. | ## Solution.
From the condition we have
$$
\begin{aligned}
& \frac{4 \tan x}{1+\tan^{2} x}+3 \tan x-5=0 \Leftrightarrow 3 \tan^{3} x-5 \tan^{2} x+7 \tan x-5=0 \\
& 3 \tan^{3} x-3 \tan^{2} x-2 \tan^{2} x+2 \tan x+5 \tan x-5=0 \\
& 3 \tan^{2} x(\tan x-1)-2 \tan x(\tan x-1)+5(\tan x-1)=0 \\
& (\tan x-1)\left(3 \tan^{2} ... | \frac{\pi}{4}(4k+1),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,570 |
8.282. $5 \sin ^{4} 2 z-4 \sin ^{2} 2 z \cos ^{2} 2 z-\cos ^{4} 2 z+4 \cos 4 z=0$. | Solution.
Rewrite the equation as
$5\left(\sin ^{2} 2 z\right)^{2}-\sin ^{2} 4 z-\left(\cos ^{2} 2 z\right)^{2}+4 \cos 4 z=0 \Leftrightarrow$
$\Leftrightarrow 5 \cdot\left(\frac{1}{2}(1-\cos 4 z)\right)^{2}-\left(1-\cos ^{2} 4 z\right)-\left(\frac{1}{2}(1-\cos 4 z)\right)^{2}+4 \cos 4 z=0 \Leftrightarrow$
$\Leftrig... | z_{1}=\frac{\pi}{8}(2k+1);z_{2}=\frac{\pi}{6}(3n\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,571 |
8.284. $\cos ^{6} x+\sin ^{6} x-\cos ^{2} 2 x=\frac{1}{16}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
8.284. $\cos ^{6} x+\sin ^{6} x-\cos ^{2} 2 x=\frac{1}{16}$. | ## Solution.
We have
$$
\begin{aligned}
& \left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{4} x-\cos ^{2} x \sin ^{2} x+\sin ^{4} x\right)-\cos ^{2} 2 x-\frac{1}{16}=0 \Leftrightarrow \\
& \Leftrightarrow\left(\cos ^{2} x+\sin ^{2} x\right)^{2}-3 \cos ^{2} x \sin ^{2} x-\cos ^{2} 2 x-\frac{1}{16}=0 \Leftrightarrow \\... | \frac{\pi}{12}(6k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,573 |
8.285. $\frac{1}{\sin ^{2} 2 x}+\tan x-\cot x-4=0$. | Solution.
Domain of definition: $\sin 2 x \neq 0$.
Rewrite the equation as
$\frac{1}{\sin ^{2} 2 x}+\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right)-4=0 \Leftrightarrow \frac{1}{\sin ^{2} 2 x}+\frac{\sin ^{2} x-\cos ^{2} x}{\sin x \cos x}-4=0 \Leftrightarrow$ $\frac{1}{\sin ^{2} 2 x}-\frac{2 \cos 2 x}{\sin 2... | x_{1}=\frac{3\pi}{8}+\frac{\pik}{2};x_{2}=\frac{1}{2}\operatorname{arcctg}3+\frac{\pin}{2},k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,574 |
8.286. $\operatorname{tg} 5 z-\operatorname{tg} 3 z-2 \operatorname{tg} 2 z=0$.
8.286. $\tan 5z - \tan 3z - 2 \tan 2z = 0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 5 z \neq 0, \\ \cos 3 z \neq 0, \\ \cos 2 z \neq 0 .\end{array}\right.$
By the formula $\operatorname{tg} \alpha-\operatorname{tg} \beta=\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta}$ we have
$\frac{\sin 2 z}{\cos 5 z \cos 3 z}-\frac{2 \sin 2 z}{\cos ... | z_{1}=\pik;z_{2}=\frac{\pi}{16}(2n+1),kn\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,575 |
8.287. $\cos 2 x+\cos \frac{3 x}{4}-2=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.287. $\cos 2 x+\cos \frac{3 x}{4}-2=0$. | Solution.
$\cos 2 x+\cos \frac{3 x}{4}=2 \Leftrightarrow\left\{\begin{array}{l}\cos 2 x=1, \\ \cos \frac{3 x}{4}=1\end{array} \Leftrightarrow\left\{\begin{array}{l}2 x=2 \pi k, \\ \frac{3 x}{4}=2 \pi l,\end{array} \Leftrightarrow\right.\right.$
$\Leftrightarrow\left\{\begin{array}{l}x=\pi k, k \in Z, \\ x=\frac{8 \pi ... | 8\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,576 |
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