problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
7.083. $x^{\log _{4} x-2}=2^{3\left(\log _{4} x-1\right)}$.
7.083. $x^{\log _{4} x-2}=2^{3\left(\log _{4} x-1\right)}$. | Solution.
Domain of definition: $0<x \neq 1$.
Taking the logarithm of both sides of the equation with base 4, we have
$$
\begin{aligned}
& \log _{4} x^{\log _{4} x-2}=\log _{4} 2^{3\left(\log _{4} x-1\right)},\left(\log _{4} x-2\right) \log _{4} x=3\left(\log _{4} x-1\right) \log _{4} 2, \\
& \log _{4}^{2} x-2 \log ... | 2;64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,851 |
7.084. $\frac{2^{x}+10}{4}=\frac{9}{2^{x-2}}$.
7.084. $\frac{2^{x}+10}{4}=\frac{9}{2^{x-2}}$. | ## Solution.
From the condition
$$
\frac{2^{x}+10}{4}=\frac{9}{2^{x} \cdot 2^{-2}}, \frac{2^{x}+10}{4}=\frac{36}{2^{x}}, 2^{2 x}+10 \cdot 2^{x}-144=0 .
$$
Solving this equation as a quadratic in terms of $2^{x}$, we find $2^{x}=-18, \varnothing$, or $2^{x}=8$, from which $x=3$.
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,852 |
7.085. $10^{1+x^{2}}-10^{1-x^{2}}=99$. | Solution.
We have $10 \cdot 10^{x^{2}}-\frac{10}{10^{x^{2}}}-99=0 \Rightarrow 10 \cdot 10^{2 x^{2}}-99 \cdot 10^{x^{2}}-10=0$. Solving this equation as a quadratic in terms of $10^{x^{2}}$, we get $10^{x^{2}}=-\frac{1}{10}, \varnothing$, or $10^{x^{2}}=10$, from which $x^{2}=1, x_{1,2}= \pm 1$.
Answer: $-1 ; 1$. | -1;1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,853 |
7.088. $3 \cdot 4^{\log _{x} 2}-46 \cdot 2^{\log _{x} 2-1}=8$. | Solution.
Domain of definition: $0<x \neq 1$.
We have $3 \cdot 2^{2 \log _{x} 2}-23 \cdot 2^{\log _{x} 2}-8=0$. Solving the equation as a quadratic in terms of $2^{\log _{x} 2}$, we find $2^{\log _{x} 2}=-\frac{1}{3}, \varnothing$; or $2^{\log _{x} 2}=8$, from which $\log _{x} 2=3, x=\sqrt[3]{2}$.
Answer: $\sqrt[3]{... | \sqrt[3]{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,854 |
7.090. $27^{\lg x}-7 \cdot 9^{\lg x}-21 \cdot 3^{\lg x}+27=0$. | ## Solution.
Domain of definition: $x>0$.
We have
$$
\begin{aligned}
& 3^{3 \lg x}-7 \cdot 3^{2 \lg x}-21 \cdot 3^{\lg x}+27=0,\left(3^{3 \lg x}+27\right)-7 \cdot 3^{\lg x}\left(3^{\lg x}+3\right)=0 \\
& \left(3^{\lg x}+3\right)\left(3^{2 \lg x}-3 \cdot 3^{\lg x}+9\right)-7 \cdot 3^{\lg x}\left(3^{\lg x}+3\right)=0 ... | 1;100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,856 |
7.091. $\log _{2}\left(4 \cdot 3^{x}-6\right)-\log _{2}\left(9^{x}-6\right)=1$. | ## Solution.
oDZ: $\left\{\begin{array}{l}4 \cdot 3^{x}-6>0 \\ 9^{x}-6>0\end{array}\right.$
We have $\log _{2} \frac{4 \cdot 3^{x}-6}{3^{2 x}-6}=1, \frac{4 \cdot 3^{x}-6}{3^{2 x}-6}=2 \Rightarrow 3^{2 x}-2 \cdot 3^{x}-3=0$. Solving it as a quadratic equation in terms of $3^{x}$, we find $3^{x}=-1, \varnothing$; or $3... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,857 |
7.095. $\sqrt{\log _{a} x}+\sqrt{\log _{x} a}=\frac{10}{3}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\log _{a} x \geq 0, \\ 0<a \neq 1, \\ 0<x \neq 1 .\end{array}\right.$
From the condition
$$
\sqrt{\log _{a} x}+\frac{1}{\sqrt{\log _{a} x}}-\frac{10}{3}=0 \Rightarrow 3\left(\sqrt{\log _{a} x}\right)^{2}-10 \sqrt{\log _{a} x}+3=0
$$
Solving this equation as a ... | \sqrt[9]{};^{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,858 |
7.096. $\lg \left(3 x^{2}+12 x+19\right)-\lg (3 x+4)=1$. | ## Solution.
Domain of definition: $3 x+4>0, x>-\frac{4}{3}$.
We have $\lg \frac{3 x^{2}+12 x+19}{3 x+4}=1, \frac{3 x^{2}+12 x+19}{3 x+4}=10,3 x^{2}-18 x-21=0$
for $3 x+4 \neq 0$. From this, $x_{1}=-1, x_{2}=7$.
Answer: $-1 ; 7$. | -1;7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,859 |
7.097. $\log _{3}(x-3)^{2}+\log _{3}|x-3|=3$. | ## Solution.
Domain of definition: $x-3 \neq 0, x \neq 3$.
From the condition
$2 \log _{3}|x-3|+\log _{3}|x-3|=3, 3 \log _{3}|x-3|=3, \log _{3}|x-3|=1$,
hence $|x-3|=3$. Then $(x-3)_{1}=-3$ or $(x-3)_{2}=3$. From this, $x_{1}=0$, $x_{2}=6$.
Answer: $0 ; 6$. | 0;6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,860 |
7.098. $\lg \sqrt{x-3}+\lg \sqrt{x+3}=2-0.5 \lg 625$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x-3>0, \\ x+3>0\end{array} x>3\right.$.
## We have
$\lg \sqrt{x-3}+\lg \sqrt{x+3}=\lg 100-\lg 25, \lg \sqrt{x^{2}-9}=\lg 4, \sqrt{x^{2}-9}=4$, from which $x^{2}=25, x_{1}=-5, x_{2}=5, x_{1}=-5$ does not satisfy the domain of definition.
Answer: 5 . | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,861 |
7.099. $\lg (3-x)-\frac{1}{3} \lg \left(27-x^{3}\right)=0$. | ## Solution.
Domain of definition: $3-x>0, x<3$.
## Rewrite the equation as
$$
3 \lg (3-x)=\lg \left(27-x^{3}\right), \lg (3-x)^{3}=\lg \left(27-x^{3}\right)
$$
Then $(3-x)^{3}=27-x^{3} \Rightarrow x^{2}-9 x=0$, from which $x_{1}=0, x_{2}=9 ; x_{2}=9$ does not satisfy the domain of definition.
Answer: 0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,862 |
7.100. $2 \lg x-\lg 4=-\lg \left(5-x^{2}\right)$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ 5-x^{2}>0,\end{array} \quad 0<x<\sqrt{5}\right.$.
From the condition
$$
\begin{aligned}
& \lg x^{2}+\lg \left(5-x^{2}\right)=\lg 4, \lg \left(x^{2}\left(5-x^{2}\right)\right)=\lg 4, x^{2}\left(5-x^{2}\right)=4 \\
& x^{4}-5 x^{2}+4=0
\end{aligned}
$$
... | 1;2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,863 |
7.101. $\lg 8-\lg \sqrt{x+6}=\lg 16-\lg (x-2)$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x+6>0, \\ x-2>0,\end{array} \quad x>2\right.$.
We have
$$
\lg \frac{8}{\sqrt{x+6}}=\lg \frac{16}{x-2}, \quad \frac{8}{\sqrt{x+6}}=\frac{16}{x-2}, \quad 2 \sqrt{x+6}=x-2, x^{2}-8 x-20=0
$$
from which $x_{1}=10, x_{2}=-2 ; x_{2}=-2$ does not satisfy the domain o... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,864 |
7.102. $2 \lg \sqrt{4-x}+\lg (6-x)=1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}4-x>0, \\ 6-x>0,\end{array} \quad x<4\right.$.
Rewrite the equation as
$$
\lg (4-x)+\lg (6-x)=1, \quad \lg (4-x)(6-x)=1
$$
from which $(4-x)(6-x)=10, x^{2}+10 x-14=0$. Therefore, $x_{1}=5-\sqrt{11}$, $x_{2}=5+\sqrt{11} ; x_{2}=5+\sqrt{11}$ does not satisfy ... | 5-\sqrt{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,865 |
7.103. $\frac{\lg (2 x-19)-\lg (3 x-20)}{\lg x}=-1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}2 x-19>0, \\ 3 x-20>0,\end{array} \quad x>\frac{19}{2}\right.$.
## From the condition
$$
\begin{aligned}
& \lg (2 x-19)-\lg (3 x-20)=-\lg x, \lg (2 x-19)+\lg x=\lg (3 x-20) \\
& x(2 x-19)=3 x-20, x^{2}-11 x+10=0 .
\end{aligned}
$$
From here $x_{1}=10, x_{2}... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,866 |
7.104. $\frac{\lg x^{2}}{\lg (6 x-5)}=1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ 6 x-5>0,\end{array}, \frac{5}{6}<x \neq 1\right.$.
We have $\lg x^{2}=\lg (6 x-5)$, from which $x^{2}=6 x-5, x^{2}-6 x+5=0$, hence $x_{1}=5$ and $x_{2}=1 ; x_{2}=1$ does not satisfy the domain of definition.
Answer: 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,867 |
7.106. $\log _{x} \sqrt{2}-\log _{x}^{2} \sqrt{2}=\log _{3} 27-\log _{x}(2 x)$. | Solution.
Domain of definition: $0<x \neq 1$.
Rewrite the equation as
$$
\frac{1}{2} \log _{x} 2-\frac{1}{4} \log _{x}^{2} 2=3-\log _{x} 2-1, \quad \log _{x}^{2} 2-6 \log _{x} 2+8=0
$$
Solving this equation as a quadratic in terms of $\log _{x} 2$, we find $\log _{x} 2=2, \log _{x} 2=4$, hence $x^{2}=2$ or $x^{4}=2... | \sqrt[4]{2};\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,868 |
7.107. $\left(\log _{2} x-3\right) \log _{2} x+2\left(\log _{2} x+1\right) \log _{2} \sqrt[3]{2}=0$. | ## Solution.
Domain of definition: $x>0$.
From the condition
$$
\begin{aligned}
& \log _{2} \sqrt[3]{2}=\log _{2} 2^{1 / 3}=\frac{1}{3}, \log _{2}^{2} x-3 \log _{2} x+\frac{2}{3} \log _{2} x+\frac{2}{3}=0 \\
& 3 \log _{2}^{2} x-7 \log _{2} x+2=0
\end{aligned}
$$
Solving the equation as a quadratic in terms of $\log... | \sqrt[3]{2};4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,869 |
7.108. $0,1 \log _{2}^{4}(x-4)-1,3 \log _{2}^{2}(x-4)+3,6=0$.
7.108. $0.1 \log _{2}^{4}(x-4)-1.3 \log _{2}^{2}(x-4)+3.6=0$. | ## Solution.
Domain of definition: $x-4>0, x>4$.
Solving this equation as a biquadratic equation in terms of $\log _{2}(x-4)$, we have $\left(\log _{2}(x-4)\right)_{1}=-2 ;\left(\log _{2}(x-4)\right)_{2}=2 ;\left(\log _{2}(x-4)\right)_{3}=-3 ;\left(\log _{2}(x-4)\right)_{4}=3$, from which $x_{1}=\frac{17}{4}, x_{2}=8... | \frac{17}{4};\frac{33}{8};8;12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,870 |
7.109. $5^{2 x-1}+2^{2 x}-5^{2 x}+2^{2 x+2}=0$. | Solution.
Write the equation as
$$
\begin{aligned}
& \frac{5^{2 x}}{5}-5^{2 x}=-2^{2 x}-4 \cdot 2^{2 x},-\frac{4}{5} \cdot 5^{2 x}=-5 \cdot 2^{2 x} \\
& \left(\frac{5}{2}\right)^{2 x}=\left(\frac{5}{2}\right)^{2}, x=1
\end{aligned}
$$
Answer: 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,871 |
7.110. $\log _{2}\left(9-2^{x}\right)=10^{\lg (3-x)}$.
7.110. $\log _{2}\left(9-2^{x}\right)=10^{\lg (3-x)}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}9-2^{x}>0, \\ 3-x>0,\end{array} \quad x<3\right.$.
We have $\log _{2}\left(9-2^{x}\right)=3-x, 9-2^{x}=2^{3-x}, 2^{2 x}-9 \cdot 2^{x}+8=0$. Solving this equation as a quadratic in terms of $2^{x}$, we get $\left(2^{x}\right)=1$ or $\left(2^{x}\right)_{2}=8$, ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,872 |
7.111. $\frac{1}{3} \lg \left(271+3^{2 \sqrt{x}}\right)+\lg 10=2$. | ## Solution.
Domain: $x \geq 0$.
From the condition $\frac{1}{3} \lg \left(271+3^{2 \sqrt{x}}\right)+1=2, \lg \left(271+3^{2 \sqrt{x}}\right)=3$. Then $271+3^{2 \sqrt{x}}=$ $=1000,3^{2 \sqrt{x}}=3^{6}$, from which $\sqrt{x}=3, x=9$.
Answer: 9. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,873 |
7.112. $\left.(\sqrt[5]{27})^{\frac{x}{4}-\sqrt{\frac{x}{3}}}\right)^{\frac{x}{4}+\sqrt{\frac{x}{3}}}=\sqrt[4]{3^{7}}$. | ## Solution.
Domain of definition: $x \geq 0$.
Rewrite the equation as $3^{\frac{3}{5}\left(\frac{x}{4}-\sqrt{\frac{x}{3}}\right)\left(\frac{x}{4}+\sqrt{\frac{x}{3}}\right)}=3^{\frac{7}{4}}$. Then
$$
\frac{3}{5}\left(\frac{x}{4}-\sqrt{\frac{x}{3}}\right)\left(\frac{x}{4}+\sqrt{\frac{x}{3}}\right)=\frac{7}{4}, 3 x^{2... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,874 |
7.113. $x^{\lg x}=1000 x^{2}$. | Solution.
Domain of definition: $0<x \neq 1$.
Taking the logarithm of both sides of the equation with base 10, we get $\lg x^{\lg x}=\lg 1000 x^{2}, \lg x \lg x=\lg 1000+\lg x^{2}, \lg ^{2} x-2 \lg x-3=0$. Solving this equation as a quadratic equation in terms of $\lg x$, we get $(\lg x)_{1}=-1$ or $(\lg x)_{2}=3$, f... | 0.1;1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,875 |
7.114. $\lg (x(x+9))+\lg \frac{x+9}{x}=0$. | ## Solution.
Domain of definition: $x(x+9)>0, x \in(-\infty ;-9) \cup(0 ; \infty)$.
We have $\lg \frac{x(x+9)(x+9)}{x}=0$, from which $(x+9)^{2}=1$. Then $(x+9)_{1}=-1$, $x_{1}=-10$ or $(x+9)_{2}=1, x_{2}=-8 ; x_{2}=-8$ does not fit the domain of definition.
Answer: -10. | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,876 |
7.115. $\lg ^{2}(100 x)+\lg ^{2}(10 x)=14+\lg \frac{1}{x}$. | ## Solution.
Domain: $x>0$.
Taking logarithms, we have
$$
(\lg 100+\lg x)^{2}+(\lg 10+\lg x)^{2}=14-\lg x, \quad 2 \lg ^{2} x+7 \lg x-9=0
$$
Solving this equation as a quadratic in $\lg x$, we get $(\lg x)_{1}=-\frac{9}{2}$ or $(\lg x)_{2}=1$, from which $x_{1}=10^{-9 / 2}, x_{2}=10$.
Answer: $\quad 10^{-9 / 2} ; ... | 10^{-9/2};10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,877 |
7.116. $1+2 \log _{x} 2 \cdot \log _{4}(10-x)=\frac{2}{\log _{4} x}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}0<x \neq 1, \\ x<10 .\end{array}\right.$
By changing to base 2, we have
$1+\frac{\log _{2}(10-x)}{\log _{2} x}=\frac{4}{\log _{2} x}, \log _{2} x+\log _{2}(10-x)=4, \quad \log _{2} x(10-x)=4 \Rightarrow$
$\Rightarrow x^{2}-10 x+16=0$,
from which $x_{1}=2, x_{... | 2;8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,878 |
7.117. $2^{\log _{3} x^{2}} \cdot 5^{\log _{3} x}=400$. | Solution.
Domain: $x>0$.
From the condition $4^{\log _{3} x} \cdot 5^{\log _{3} x}=400, 20^{\log _{3} x}=20^{2}$, hence $\log _{3} x=2$, $x=9$.
Answer: 9. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,879 |
7.121. $\log _{2}\left(4^{x}+4\right)=x+\log _{2}\left(2^{x+1}-3\right)$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.121. $\log _{2}\left(4^{x}+4\right)=x+\log _{2}\left(2^{x+1}-3\right)$ | ## Solution.
Domain of definition: $2^{x+1}-3>0$.
Rewrite the equation as
$$
\begin{aligned}
& \log _{2}\left(2^{2 x}+4\right)-\log _{2}\left(2 \cdot 2^{x}-3\right)=x, \quad \log _{2} \frac{2^{2 x}+4}{2 \cdot 2^{x}-3}=x \\
& \frac{2^{2 x}+4}{2 \cdot 2^{x}-3}=2^{x}, 2^{2 x}-3 \cdot 2^{x}-4=0 .
\end{aligned}
$$
Solvi... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,880 |
7.122. $\sqrt[3]{27^{5 \sqrt{x}}}=3^{x(\sqrt{x}-4)}$.
7.122. $\sqrt[3]{27^{5 \sqrt{x}}}=3^{x(\sqrt{x}-4)}$. | Solution.
Domain of definition: $x \geq 0$.
We have $3^{5 \sqrt{x}}=3^{x(\sqrt{x}-4)} \Rightarrow 5 \sqrt{x}=x(\sqrt{x}-4) \quad \sqrt{x}=0, x_{1}=0, \quad$ or $(\sqrt{x})^{2}-4 \sqrt{x}-5=0$.
Solving this equation as a quadratic equation in terms of $\sqrt{x}$, we get $\sqrt{x}=-1, \varnothing$; or $\sqrt{x}=5, x=2... | 0;25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,881 |
7.123. $\log _{6} \sqrt[7]{3^{x(15-x)}}+8 \log _{6} 2=8$. | ## Solution.
From the condition $\log _{6} 3^{x(15-x) / 7}+\log _{6} 2^{8}=8, \quad \log _{6}\left(3^{x(15-x) / 7} \cdot 2^{8}\right)=8$. Therefore, $3^{x(15-x) / 7} \cdot 2^{8}=6^{8}, 3^{x(15-x) / 7}=3^{8}$. Then $\frac{x(15-x)}{7}=8, x^{2}-15 x+56=0$, from which $x_{1}=7, x_{2}=8$.
Answer: $7 ; 8$. | 7;8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,882 |
7.124. $\log _{5}\left(4^{x}+144\right)-4 \log _{5} 2=1+\log _{5}\left(2^{x-2}+1\right)$ | ## Solution.
We have
$$
\begin{aligned}
& \log _{5} \frac{2^{2 x}+144}{16}=\log _{5} 5\left(\frac{2^{x}}{4}+1\right) \frac{2^{2 x}+144}{16}=\frac{5\left(2^{x}+4\right)}{4} \\
& 2^{2 x}-20 \cdot 2^{x}+64=0
\end{aligned}
$$
- Solving this equation as a quadratic in terms of $2^{x}$, we get $\left(2^{x}\right)_{1}=4$ o... | 2;4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,883 |
7.126. $\log _{x} 9+\log _{x^{2}} 729=10$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ x \neq \pm 1,\end{array} \quad 0<x \neq 1\right.$.
We have $\log _{x} 9+\frac{3}{2} \log _{x} 9=10, \log _{x} 9=4$, from which $x^{4}=9, x=\sqrt{3}$, $x=-\sqrt{3}$ does not fit the domain of definition.
Answer: $\sqrt{3}$. | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,885 |
7.127. $\log _{2}\left(25^{x+3}-1\right)=2+\log _{2}\left(5^{x+3}+1\right)$ | Solution.
Domain of definition: $25^{x+3}-1>0, 25^{x+3}>25^{0}, x>-3$.
From the condition
$\log _{2}\left(25^{3} \cdot 25^{x}-1\right)=\log _{2} 4\left(5^{3} \cdot 5^{x}+1\right) 25^{3} \cdot 5^{2 x}-1=4 \cdot 5^{3} \cdot 5^{x}+4$, $3125 \cdot 5^{2 x}-100 \cdot 5^{x}-1=0$,
from which, solving this equation as a qua... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,886 |
7.128. $\left\{\begin{array}{l}\log _{y} x+\log _{x} y=2, \\ x^{2}-y=20\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
\log _{y} x+\log _{x} y=2, \\
x^{2}-y=20
\end{array}\right.
\] | Solution.
Domain of definition: $\left\{\begin{array}{l}0<x \neq 1, \\ 0<y \neq 1 .\end{array}\right.$
From the first equation, we have:
$\log _{y} x+\frac{1}{\log _{y} x}-2=0, \quad \log _{y}^{2} x-2 \log _{y} x+1=0, \quad\left(\log _{y} x-1\right)^{2}=0 ;$
from which $\log _{y} x=1, x=y$. From the second equation ... | (5;5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,887 |
7.129. $\left\{\begin{array}{l}10^{1+\lg (x+y)}=50, \\ \lg (x-y)+\lg (x+y)=2-\lg 5 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x-y>0, \\ x+y>0 .\end{array}\right.$
## We have:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ \operatorname { l g } 1 0 ^ { 1 + \operatorname { l g } ( x + y ) } = \operatorname { l g } 5 0 , } \\
{ \operatorname { l g } ( x ^ { 2 } - y ^ { 2 } ) = \opera... | (\frac{9}{2};\frac{1}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,888 |
7.130. $\left\{\begin{array}{l}\lg \left(x^{2}+y^{2}\right)=2-\lg 5, \\ \lg (x+y)+\lg (x-y)=\lg 1.2+1 .\end{array}\right.$ | Solution.
Domain of Definition (DOD): $\left\{\begin{array}{l}x+y>0, \\ x-y>0 .\end{array}\right.$
## From the condition
$$
\left\{\begin{array} { l }
{ \operatorname { l g } ( x ^ { 2 } + y ^ { 2 } ) = \operatorname { l g } 2 0 , } \\
{ \operatorname { l g } ( x ^ { 2 } - y ^ { 2 } ) = \operatorname { l g } 1 2 }
... | (4;2)(4;-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,889 |
7.131. $\left\{\begin{array}{l}\log _{4} x+\log _{4} y=1+\log _{4} 9, \\ x+y-20=0 .\end{array}\right.$ | Solution.
Domain of Definition: $\left\{\begin{array}{l}x>0, \\ y>0 .\end{array}\right.$
We have $\left\{\begin{array}{l}x y=36, \\ x+y=20,\end{array}\right.$ from which $\left\{\begin{array}{l}x_{1}=2, \\ y_{1}=18 ;\end{array}\left\{\begin{array}{l}x_{2}=18, \\ y_{2}=2 .\end{array}\right.\right.$
Answer: $\quad(2 ;... | (2;18),(18;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,890 |
7.132. $\left\{\begin{array}{l}3^{y} \cdot 9^{x}=81 \\ \lg (y+x)^{2}-\lg x=2 \lg 3 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ y+x \neq 0 .\end{array}\right.$
From the first equation of the system $3^{y+2 x}=3^{4}, y+2 x=4, y=4-2 x$.
From the second equation of the system $\lg \frac{(y+x)^{2}}{x}=\lg 9$, hence $\frac{(y+x)^{2}}{x}=9$. Then the original system becomes
$$
\le... | (1;2),(16;-28) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,891 |
7.134. $\left\{\begin{array}{l}3^{2 x}-2^{y}=725 \\ 3^{x}-2^{y / 2}=25\end{array}\right.$
Solve the system of equations:
\[
\left\{\begin{array}{l}
3^{2 x}-2^{y}=725 \\
3^{x}-2^{y / 2}=25
\end{array}\right.
\] | ## Solution.
Rewrite the system of equations as
$$
\left\{\begin{array} { l }
{ ( 3 ^ { x } - 2 ^ { y / 2 } ) ( 3 ^ { x } + 2 ^ { y / 2 } ) = 7 2 5 , } \\
{ 3 ^ { x } - 2 ^ { y / 2 } = 2 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ 3 ^ { x } + 2 ^ { y / 2 } = 2 9 , } \\
{ 3 ^ { x } - 2 ^ { y / 2 } =... | (3;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,892 |
7.135. $\left\{\begin{array}{l}\lg \left(x^{2}+y^{2}\right)=2, \\ \log _{2} x-4=\log _{2} 3-\log _{2} y .\end{array}\right.$
The system of equations is:
7.135. $\left\{\begin{array}{l}\lg \left(x^{2}+y^{2}\right)=2, \\ \log _{2} x-4=\log _{2} 3-\log _{2} y .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0 .\end{array}\right.$
From the first equation of the system, we have $x^{2}+y^{2}=100$. From the second equation of the system, we find $\log _{2} \frac{x}{16}=\log _{2} \frac{3}{y}$, from which $\frac{x}{16}=\frac{3}{y}$, $x=\frac{48}{y}$. Further... | (8;6)(6;8) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,893 |
7.136. $\left\{\begin{array}{l}3^{2 \sqrt{x}-\sqrt{y}}=81 \\ \lg \sqrt{x y}=1+\lg 3\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
3^{2 \sqrt{x}-\sqrt{y}}=81 \\
\lg \sqrt{x y}=1+\lg 3
\end{array}\right.
\] | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0 .\end{array}\right.$
From the first equation of the system, we have $3^{2 \sqrt{x}-\sqrt{y}}=3^{4}, 2 \sqrt{x}-\sqrt{y}=4$, $\sqrt{y}=2 \sqrt{x}-4$. From the second equation of the system, we get $\sqrt{x y}=30$, $\sqrt{x} \cdot \sqrt{y}=30$. The ... | (25;36) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,894 |
7.137. $\left\{\begin{array}{l}2^{\frac{x-y}{2}}+2^{\frac{y-x}{x}}=2.5 \\ \lg (2 x-y)+1=\lg (y+2 x)+\lg 6\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}2 x-y>0, \\ y+2 x>0 .\end{array}\right.$
From the first equation of the system, we get $\left(2^{\frac{x-y}{2}}\right)^{2}-2.5 \cdot 2^{\frac{x-y}{2}}+1=0$. Solving this equation as a quadratic equation in terms of $2^{\frac{x-y}{2}}$, we find $\left(2^{\frac... | (4;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,895 |
7.138. $\left\{\begin{array}{l}x^{2 y^{2}-1}=5 \\ x^{y^{2}+2}=125 .\end{array}\right.$ | ## Solution.
Domain of definition: $\quad 0<x \neq 1$.
By logarithmizing the first and second equations of the system with base 5, we get
$$
\left\{\begin{array} { l }
{ \operatorname { l o g } _ { 5 } x ^ { 2 y ^ { 2 } - 1 } = \operatorname { l o g } _ { 5 } 5 , } \\
{ \operatorname { l o g } _ { 5 } x ^ { y ^ { 2... | (5;1)(5;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,896 |
7.140. $\left\{\begin{array}{l}4^{x+y}=2^{y-x} \\ 4^{\log _{\sqrt{2}} x}=y^{4}-5\end{array}\right.$ | ## Solution.
Domain of definition: $x>0$. From the condition
$$
\left\{\begin{array} { l }
{ 2 ^ { 2 x + 2 y } = 2 ^ { y - x } , } \\
{ 2 ^ { \log _ { 2 } x ^ { 4 } } = y ^ { 4 } - 5 }
\end{array} \Rightarrow \left\{\begin{array}{l}
2 x+2 y=y-x, \\
x^{4}=y^{4}-5
\end{array} \Rightarrow y=-3 x\right.\right.
$$
From ... | (\frac{1}{2};-\frac{3}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,897 |
7.142. $\left\{\begin{array}{l}\log _{2} x+\log _{4} y=4, \\ \log _{4} x+\log _{2} y=5\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
\log _{2} x+\log _{4} y=4, \\
\log _{4} x+\log _{2} y=5
\end{array}\right.
\] | Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0 .\end{array}\right.$
Let's switch to base 2. We have
$\left\{\begin{array}{l}\log _{2} x+\frac{1}{2} \log _{2} y=4, \\ \frac{1}{2} \log _{2} x+\log _{2} y=5\end{array} \Leftrightarrow\left\{\begin{array}{l}2 \log _{2} x+\log _{2} y=8, \\ \log _{2} x... | (4;16) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,898 |
7.143. $\left\{\begin{array}{l}2^{\frac{x+y}{3}}+2^{\frac{x+y}{6}}=6 \\ x^{2}+5 y^{2}=6 x y\end{array}\right.$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.143. $\left\{\begin{array}{l}2^{\frac{x+y}{3}}+2^{\frac{x+y}{6}}=6 \\ ... | ## Solution.
From the condition $\left(2^{\frac{x+y}{6}}\right)^{2}+2^{\frac{x+y}{6}}-6=0$. Solving this equation as a quadratic equation in terms of $2^{\frac{x+y}{6}}$, we have $2^{\frac{x+y}{6}}=-3, \varnothing$; or $2^{\frac{x+y}{6}}=2$, from which $\frac{x+y}{6}=1, x+y=6$.
From the second equation of the system ... | (3;3),(5;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,899 |
7.144. $\left\{\begin{array}{l}2^{x} \cdot 3^{y}=6 \\ 3^{x} \cdot 4^{y}=12 .\end{array}\right.$ | ## Solution.
Dividing the second equation of the given system by the first, we get $\frac{3^{x} \cdot 4^{y}}{2^{x} \cdot 3^{y}}=\frac{12}{6}, \quad \frac{3^{x-y}}{2^{x-2 y}}=2, \quad 3^{x-y}=2^{1+x-2 y}$.
This equality is possible when
$$
\left\{\begin{array}{l}
x-y=0, \\
1+x-2 y=0
\end{array} \Rightarrow x=y, \quad... | (1;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,900 |
7.145. $\left\{\begin{array}{l}y=1+\log _{4} x, \\ x^{y}=4^{6}\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
y=1+\log _{4} x, \\
x^{y}=4^{6}
\end{array}\right.
\] | ## Solution.
Domain of definition: $0<x \neq 1$.
Taking the logarithm of the second equation of the system with base 4, we have $\log _{4} x^{y}=\log _{4} 4^{6}, y \log _{4} x=6$. Therefore,
$$
\left\{\begin{array}{l}
y=1+\log _{4} x, \\
y \log _{4} x=6
\end{array} \Rightarrow\left(1+\log _{4} x\right) \log _{4} x=6... | (\frac{1}{64};-2),(16;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,901 |
7.146. $\left\{\begin{array}{l}\log _{\sqrt{x}}(x y)=8 \\ \log _{3} \log _{1 / 9} \frac{x}{y}=0 .\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
\log _{\sqrt{x}}(x y)=8 \\
\log _{3} \log _{1 / 9} \frac{x}{y}=0 .
\end{array}\right.
\] | Solution.
Odz: $\left\{\begin{array}{l}00, \\ \log _{1 / 9} \frac{x}{y}>0 \Rightarrow 0<\frac{x}{y}<1 .\end{array}\right.$
From the first equation of the system $x y=x^{4}$ or considering the domain of definition (ODZ) $y=x^{3}$. From the second equation we have $\log _{1 / 9} \frac{x}{y}=1, \frac{x}{y}=\frac{1}{9}$.... | (3;27) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,902 |
7.147. $\left\{\begin{array}{l}\log _{x y}(x-y)=1, \\ \log _{x y}(x+y)=0 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x-y>0, \\ x+y>0, \\ 0<x y \neq 1 .\end{array}\right.$
## We have
$\left\{\begin{array}{l}x-y=x y, \\ x+y=1\end{array} \Rightarrow y=1-x, \quad x-(1-x)-x(1-x)=0, x^{2}+x-1=0\right.$, from which $x_{1}=\frac{-1-\sqrt{5}}{2}, x_{2}=\frac{-1+\sqrt{5}}{2}, y_{1}=... | (\frac{-1+\sqrt{5}}{2};\frac{3-\sqrt{5}}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,903 |
7.148. $\left\{\begin{array}{l}(x+y) \cdot 2^{y-2 x}=6.25, \\ (x+y) \frac{1}{2 x-y}=5 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}0<x+y \neq 1, \\ 2 x-y \neq 0 .\end{array}\right.$
Taking the logarithm of both equations with base 10, we have
$$
\left\{\begin{array} { l }
{ \operatorname { l g } ( x + y ) \cdot 2 ^ { y - 2 x } = \operatorname { l g } ( \frac { 5 } { 2 } ) ^ { 2 } , } \... | (9;16) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,904 |
7.149. $\left\{\begin{array}{l}8^{\log _{9}(x-4 y)}=1, \\ 4^{x-2 y}-7 \cdot 2^{x-2 y}=8 .\end{array}\right.$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,905 | ||
8.001. $\cos 3 x-\sin x=\sqrt{3}(\cos x-\sin 3 x)$
8.001. $\cos 3x - \sin x = \sqrt{3}(\cos x - \sin 3x)$ | ## Solution.
$\cos 3 x-\sqrt{3} \cos x=\sin x-\sqrt{3} \sin 3 x, \Leftrightarrow \frac{1}{2} \cos 3 x+\frac{\sqrt{3}}{2} \sin 3 x=$ $=\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x, \Leftrightarrow \cos 3 x \cos \frac{\pi}{3}+\sin 3 x \sin \frac{\pi}{3}=$ $=\cos \frac{\pi}{6} \cos x+\sin \frac{\pi}{6} \sin x, \Leftright... | x_{1}=\frac{\pi}{8}(4k+1),x_{2}=\frac{\pi}{12}(12n+1),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,906 |
8.002. $7+4 \sin x \cos x+1.5(\tan x+\cot x)=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos x \neq 0 .\end{array}\right.$
$$
\begin{aligned}
& 7+4 \sin x \cos x+1.5\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 7+4 \sin x \cos x+1.5\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \co... | (-1)^{k+1}\frac{\pi}{12}+\frac{\pik}{2},\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,907 |
8.003. $\frac{4 \operatorname{ctg} x}{1+\operatorname{ctg}^{2} x}+\sin ^{2} 2 x+1=0$.
8.003. $\frac{4 \cot x}{1+\cot^{2} x}+\sin ^{2} 2 x+1=0$. | ## Solution.
Domain of definition: $\sin x \neq 0$.
## From the condition
$$
\begin{aligned}
& \frac{\frac{4 \cos x}{\sin x}}{1+\frac{\cos ^{2} x}{\sin ^{2} x}}+\sin ^{2} 2 x+1=0, \Leftrightarrow \sin ^{2} 2 x+2 \sin 2 x+1=0 \Leftrightarrow \\
& \Leftrightarrow(\sin 2 x+1)^{2}=0, \sin 2 x=-1
\end{aligned}
$$
Then $... | \frac{\pi}{4}(4k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,908 |
8.004. $\frac{\sin ^{2} 2 x-4 \sin ^{2} x}{\sin ^{2} 2 x+4 \sin ^{2} x-4}+1=2 \tan^{2} x$. | ## Solution.
Domain of definition: $\cos x \neq 0$.
We have
$$
\begin{aligned}
& \frac{\sin ^{2} 2 x-4 \sin ^{2} x+\sin ^{2} 2 x+4 \sin ^{2} x-4}{\sin ^{2} 2 x+4 \sin ^{2} x-4}=2 \operatorname{tg}^{2} x \Leftrightarrow \\
& \Leftrightarrow \frac{2 \sin ^{2} 2 x-4}{\sin ^{2} 2 x+4 \sin ^{2} x-4}=2 \operatorname{tg}^{... | \frac{\pi}{4}(2+1)\quad\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,909 |
8.005. $\sin z \sin \left(60^{\circ}-z\right) \sin \left(60^{\circ}+z\right)=\frac{1}{8}$. | ## Solution.
From the condition
$$
\begin{aligned}
& \sin z\left(2 \sin \left(60^{\circ}-z\right) \sin \left(60^{\circ}+z\right)\right)=\frac{1}{4}, \Leftrightarrow \sin z\left(\cos 2 z-\cos 120^{\circ}\right)=\frac{1}{4} \Leftrightarrow \\
& \Leftrightarrow 2 \sin z \cos 2 z+\sin z=\frac{1}{2}, \Leftrightarrow-\sin ... | (-1)^{k}\cdot10+60k,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,910 |
8.007. $\operatorname{tg} 3 t-\operatorname{tg} t-4 \sin t=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.007. $\tan 3 t-\tan t-4 \sin t=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos 3 t \neq 0 \\ \cos t \neq 0\end{array}\right.$
Using the formula $\operatorname{tg} \alpha-\operatorname{tg} \beta=\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta}$, rewrite the equation as $\frac{\sin 2 t}{\cos 3 t \cos t}-4 \sin t=0, \frac{2 \sin t \cos... | t_{1}=\pik;t_{2}=\\frac{\pi}{9}+\frac{2\pin}{3},\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,912 |
8.009. $\operatorname{ctg} t-\sin t=2 \sin ^{2} \frac{t}{2}$. | ## Solution.
Domain of definition: $\sin t \neq 0$.
$$
\begin{aligned}
& \frac{\cos t}{\sin t}-\sin t=1-\cos t \Rightarrow \cos t-\sin ^{2} t=\sin t-\sin t \cos t \\
& (\cos t+\sin t \cos t)-\left(\sin ^{2} t+\sin t\right)=0, \cos t(1+\sin t)-\sin t(1+\sin t)=0 \\
& (1+\sin t)(\cos t-\sin t)=0
\end{aligned}
$$
From ... | t_{1}=\frac{\pi}{2}(4k-1);t_{2}=\frac{\pi}{4}(4n+1),\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,914 |
8.010. $8 \cos z \cos \left(60^{\circ}-z\right) \cos \left(60^{\circ}+z\right)+1=0$. | ## Solution.
We have
$4 \cos z\left(\cos 2 z+\cos 120^{\circ}\right)+1=0, \quad 4 \cos z \cos 2 z-2 \cos z+1=0 \Leftrightarrow$ $\Leftrightarrow 2 \cos z+2 \cos 3 z-2 \cos z+1=0, \quad \cos 3 z=-\frac{1}{2}, \quad 3 z= \pm \frac{2}{3} \pi+2 \pi k$, $z= \pm \frac{2}{9} \pi+\frac{2 \pi k}{3}, k \in Z$.
Answer: $z= \pm... | \\frac{2}{9}\pi+\frac{2\pik}{3},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,915 |
8.012. $\sin x \cos 2 x+\cos x \cos 4 x=\sin \left(\frac{\pi}{4}+2 x\right) \sin \left(\frac{\pi}{4}-3 x\right)$ | ## Solution.
Let's rewrite the given equation as
$$
\begin{aligned}
& -\sin x + \sin 3x + \cos 3x + \cos 5x = \cos 5x - \cos \left(\frac{\pi}{2} - x\right) \Leftrightarrow \\
& \Leftrightarrow \sin 3x + \cos 3x = 0, \sin 3x = -\cos 3x, \operatorname{tg} 3x = -1
\end{aligned}
$$
from which $3x = -\frac{\pi}{4} + \pi ... | \frac{\pi}{12}(4n-1),\quadn\in\mathbb{Z} | Algebra | proof | Yes | Yes | olympiads | false | 50,916 |
8.013. $\sin 2 x=\cos ^{4} \frac{x}{2}-\sin ^{4} \frac{x}{2$.
8.013. $\sin 2 x=\cos ^{4} \frac{x}{2}-\sin ^{4} \frac{x}{2}$.
(Note: The original text had a minor formatting issue in the equation, which has been corrected in the translation.) | Solution.
We have
$2 \sin x \cos x-\left(\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}\right)\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)=0$,
$2 \sin x \cos x-\cos x=0 \Leftrightarrow \cos x(2 \sin x-1)=0$.
Then:
1) $\cos x=0, x_{1}=\frac{\pi}{2}+\pi n=\frac{\pi}{2}(2 n+1), n \in Z$;
or
2) $2 \sin x-1=0... | x_{1}=\frac{\pi}{2}(2n+1),\quadx_{2}=(-1)^{k}\frac{\pi}{6}+\pik,\quadn,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,917 |
8.014. $(1+\cos 4 x) \sin 2 x=\cos ^{2} 2 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.014. $(1+\cos 4 x) \sin 2 x=\cos ^{2} 2 x$. | ## Solution.
From the condition
$\left(1+1-2 \sin ^{2} 2 x\right) \sin 2 x=1-\sin ^{2} 2 x, 2 \sin ^{3} 2 x-\sin ^{2} 2 x-2 \sin 2 x+1=0$, $\sin ^{2} 2 x(2 \sin 2 x-1)-(2 \sin 2 x-1)=0, \quad(2 \sin 2 x-1)\left(\sin ^{2} 2 x-1\right)=0$.
From this, either
1) $2 \sin 2 x-1=0, \sin 2 x=\frac{1}{2}, 2 x=(-1)^{k} \frac... | x_{1}=(-1)^{k}\frac{\pi}{12}+\frac{\pik}{2};x_{2}=\frac{\pi}{4}(2n+1),\quadk,n\inZ | Algebra | proof | Yes | Yes | olympiads | false | 50,918 |
8.015. $\sin ^{2} 2 z+\sin ^{2} 3 z+\sin ^{2} 4 z+\sin ^{2} 5 z=2$. | ## Solution.
Rewrite the equation as
$$
\begin{aligned}
& \frac{1}{2}(1-\cos 4 z)+\frac{1}{2}(1-\cos 6 z)+\frac{1}{2}(1-\cos 8 z)+\frac{1}{2}(1-\cos 10 z)=2 \\
& (\cos 4 z+\cos 6 z)+(\cos 8 z+\cos 10 z)=0, \Leftrightarrow 2 \cos 5 z \cos z+2 \cos 9 z \cos z=0
\end{aligned}
$$
$2 \cos z(\cos 5 z+\cos 9 z)=0$.
Then:
... | z_{1}=\frac{\pi}{14}(2n+1),z_{2}=\frac{\pi}{4}(2+1)\quadn,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,919 |
8.016. $\operatorname{ctg}^{4} 2 z+\sin ^{-4} 2 z=25$.
8.016. $\cot^{4} 2 z+\sin ^{-4} 2 z=25$. | ## Solution.
Domain of definition: $\sin 2 z \neq 0$.
From the condition
$$
\begin{aligned}
& \frac{\cos ^{4} 2 z}{\sin ^{4} 2 z}+\frac{1}{\sin ^{4} 2 z}-25=0 \Leftrightarrow \cos ^{4} 2 z+1-25 \sin ^{4} 2 z=0 \Leftrightarrow \\
& \Leftrightarrow\left(\cos ^{2} 2 z\right)^{2}+1-25 \sin ^{4} 2 z=0, \quad\left(1-\sin ... | \frac{\pi}{12}(6k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,920 |
8.018. $\operatorname{ctg}\left(\frac{3 \pi}{2}+x\right)-\operatorname{tg}^{2} x=(\cos 2 x-1) \cos ^{-2} x$.
8.018. $\cot\left(\frac{3 \pi}{2}+x\right)-\tan^{2} x=(\cos 2 x-1) \cos^{-2} x$. | ## Solution.
Domain of definition: $\cos x \neq 0$.
## We have
$-\operatorname{tg} x-\operatorname{tg}^{2} x=\frac{\cos 2 x-1}{\cos ^{2} x}, \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}=\frac{1-\cos 2 x}{\cos ^{2} x} \Leftrightarrow$
$\Leftrightarrow \frac{\sin x \cos x+\sin ^{2} x}{\cos ^{2} x}=\frac{1-\l... | x_{1}=\pik;\quadx_{2}=\frac{\pi}{4}(4n+1)\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,921 |
8.019. $\cos \frac{x}{2} \cos \frac{3 x}{2}-\sin x \sin 3 x-\sin 2 x \sin 3 x=0$. | Solution.
Rewrite the equation as
$$
\begin{aligned}
& \cos x + \cos 2x - \cos 2x + \cos 4x - \cos x + \cos 5x = 0 \Leftrightarrow \\
& \Leftrightarrow \cos 4x + \cos 5x = 0, \Leftrightarrow 2 \cos \frac{9x}{2} \cos \frac{x}{2} = 0
\end{aligned}
$$
Then either
1) $\cos \frac{9x}{2} = 0, \quad \frac{9x}{2} = \frac{\... | \frac{\pi}{9}(2k+1)\quadk\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,922 |
8.020. $1-\sin 3 x=\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^{2}$.
8.020. $1-\sin 3x=\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^{2}$. | ## Solution.
## Given
$1-\sin 3 x=\sin ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}+\cos ^{2} \cdot \frac{x}{2}, \quad \sin 3 x-\sin x=0 \Leftrightarrow$
$\Leftrightarrow 2 \sin x \cos 2 x=0$.
Then either
1) $\sin x=0, \quad x_{1}=\pi n, \quad n \in Z$,
## or
2) $\cos 2 x=0,2 x=\frac{\pi}{2}+\pi k, \quad... | x_{1}=\pin;x_{2}=\frac{\pi}{4}(2k+1),\quadn,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,923 |
8.021. $2 \operatorname{ctg}^{2} x \cos ^{2} x+4 \cos ^{2} x-\operatorname{ctg}^{2} x-2=0$. | ## Solution.
Domain of definition: $\sin x \neq 0$.
From the condition $2 \cos ^{2} x\left(\operatorname{ctg}^{2} x+2\right)-\left(\operatorname{ctg}^{2} x+2\right)=0$, or $\left(\operatorname{ctg}^{2} x+2\right)$
$\times\left(2 \cos ^{2} x-1=0\right),\left(\operatorname{ctg}^{2} x+2\right) \cos 2 x=0$. Since $\opera... | \frac{\pi}{4}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,924 |
8.022. $2 \sin ^{3} x+2 \sin ^{2} x \cos x-\sin x \cos ^{2} x-\cos ^{3} x=0$. | Solution.
Rewrite the equation as
$$
\begin{aligned}
& 2 \sin ^{2} x(\sin x+\cos x)-\cos ^{2} x(\sin x+\cos x)=0, \Leftrightarrow \\
& \Leftrightarrow(\sin x+\cos x)\left(2 \sin ^{2} x-\cos ^{2} x\right)=0 .
\end{aligned}
$$
Then:
1) $\sin x+\cos x=0$
or
2) $2 \sin ^{2} x-\cos ^{2} x=0$.
We have:
1) $\operatorna... | x_{1}=\frac{\pi}{4}(4n-1),x_{2,3}=\\operatorname{arctg}\frac{\sqrt{2}}{2}+\pik,\quadn,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,925 |
8.023. $\sin 7 x+\sin 9 x=2\left(\cos ^{2}\left(\frac{\pi}{4}-x\right)-\cos ^{2}\left(\frac{\pi}{4}+2 x\right)\right)$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.023. $\sin 7 x+\sin 9 x=2\left(\cos ^{2}\left(\frac{\pi}{4}-x\... | ## Solution.
From the condition
$2 \sin 8 x \cos x=1+\cos \left(\frac{\pi}{2}-2 x\right)-1-\cos \left(\frac{\pi}{2}+4 x\right) \Leftrightarrow$
$\Leftrightarrow 2 \sin 8 x \cos x=\sin 2 x+\sin 4 x \Leftrightarrow$
$\Leftrightarrow 2 \sin 8 x \cos x-2 \sin 3 x \cos x=0, \Leftrightarrow \cos x(\sin 8 x-\sin 3 x)=0$.
... | x_{1}=\frac{\pi}{2}(2k+1);x_{2}=\frac{2}{5}\pin;x_{3}=\frac{\pi}{11}(2+1),k,n,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,926 |
8.025. $\sin \left(15^{\circ}+x\right)+\sin \left(45^{\circ}-x\right)=1$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
8.025. $\sin \left(15^{\circ}+x\right)+\sin \left(45^{\circ}-x\right)=1$. | ## Solution.
We have:
$$
2 \sin \frac{15^{\circ}+x+45^{\circ}-x}{2} \cos \frac{15^{\circ}+x-45^{\circ}+x}{2}=1, \quad \cos \left(x-15^{\circ}\right)=1
$$
Then $x-15^{\circ}=360^{\circ} k, x=15^{\circ}+360^{\circ} k, k \in Z$.
Answer: $\quad x=15^{\circ}+360^{\circ} k, \quad k \in Z$. | 15+360k,\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,927 |
8.026. $\cos ^{-1} x+\operatorname{ctg} 3 x=\operatorname{ctg} \frac{3 x}{2}$.
8.026. $\cos ^{-1} x+\cot 3 x=\cot \frac{3 x}{2}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin 3 x \neq 0, \\ \sin \frac{3 x}{2} \neq 0\end{array}\right.$
## From the condition
$$
\begin{aligned}
& \frac{1}{\cos x}+\left(\operatorname{ctg} 3 x-\operatorname{ctg} \frac{3 x}{2}\right)=0 \Leftrightarrow \frac{1}{\cos x}-\frac{\sin \le... | x_{1}=\frac{\pi}{4}(4k+1);x_{2}=\frac{\pi}{8}(4n+1)\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,928 |
8.027. $\sin x \sin 3 x+\sin 4 x \sin 8 x=0$. | Solution.
Rewrite the equation as
$$
\begin{aligned}
& \frac{1}{2}(\cos (x-3 x)-\cos (x+3 x))+\frac{1}{2}(\cos (4 x-8 x)-\cos (4 x+8 x))=0 \\
& \cos 2 x-\cos 12 x=0 \Leftrightarrow 2 \sin \frac{2 x+12 x}{2} \sin \frac{12 x-2 x}{2}=0, \sin 7 x \sin 5 x=0
\end{aligned}
$$
From this:
1) $\sin 7 x=0,7 x=\pi n, \quad x_... | x_{1}=\frac{\pin}{7};x_{2}=\frac{\pik}{5},\quadn,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,929 |
8.028. $2 \operatorname{tg}^{3} x-2 \operatorname{tg}^{2} x+3 \operatorname{tg} x-3=0$.
8.028. $2 \tan^{3} x-2 \tan^{2} x+3 \tan x-3=0$. | ## Solution.
Domain of definition: $\cos x \neq 0$.
From the condition $2 \operatorname{tg}^{2} x(\operatorname{tg} x-1)+3(\operatorname{tg} x-1)=0, \quad(\operatorname{tg} x-1)\left(2 \operatorname{tg}^{2} x+3\right)=0$.
Hence $\operatorname{tg} x-1=0, \operatorname{tg} x=1, x=\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n+... | \frac{\pi}{4}(4n+1),\quadn\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,930 |
8.029. $\cos x \cos 2 x=\sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}+4 x\right)+\sin \left(\frac{3 \pi}{4}+4 x\right) \cos \left(\frac{7 \pi}{4}-5 x\right)$ | ## Solution.
Rewrite the equation as
$$
\begin{aligned}
& \frac{1}{2}(\cos (x-2 x)+\cos (x+2 x))= \\
& =\frac{1}{2}\left(\cos \left(\frac{\pi}{4}+x-\frac{\pi}{4}-4 x\right)-\cos \left(\frac{\pi}{4}+x+\frac{\pi}{4}+4 x\right)\right)+ \\
& +\frac{1}{2}\left(\sin \left(\frac{3 \pi}{4}+4 x-\frac{7 \pi}{4}+5 x\right)+\sin... | x_{1}=\frac{\pik}{2};x_{2}=\frac{\pi}{14}(2n+1),k,n\inZ | Algebra | proof | Yes | Yes | olympiads | false | 50,931 |
8.030. $2+\tan x \cot \frac{x}{2}+\cot x \tan \frac{x}{2}=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0, \\ \sin \frac{x}{2} \neq 0, \\ \cos \frac{x}{2} \neq 0 .\end{array}\right.$
Using the half-angle formulas $\operatorname{ctg} \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}, \operatorname{tg} \frac{\alpha}{2}=\frac{\sin \alph... | \frac{2\pi}{3}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,932 |
8.031. $\sin 2 x+\sin (\pi-8 x)=\sqrt{2} \cos 3 x$.
8.031. $\sin 2 x+\sin (\pi-8 x)=\sqrt{2} \cos 3 x$. | Solution.
From the condition
$\sin 2 x+\sin 8 x-\sqrt{2} \cos 3 x=0 \Leftrightarrow 2 \sin \frac{2 x+8 x}{2} \cos \frac{2 x-8 x}{2}-\sqrt{2} \cos 3 x=0$, $2 \sin 5 x \cos 3 x-\sqrt{2} \cos 3 x=0, \cos 3 x(2 \sin 5 x-\sqrt{2})=0$.
From this
1) $\cos 3 x=0, \quad 3 x=\frac{\pi}{2}+\pi k, \quad x_{1}=\frac{\pi}{6}+\fr... | x_{1}=\frac{\pi}{6}(2k+1),\quadx_{2}=(-1)^{n}\frac{\pi}{20}+\frac{\pin}{5},\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,933 |
8.032. $0.5(\cos 5 x+\cos 7 x)-\cos ^{2} 2 x+\sin ^{2} 3 x=0$. | ## Solution.
## Rewrite the equation as
$\cos \frac{5 x+7 x}{2} \cos \frac{5 x-7 x}{2}-\frac{1}{2}(1+\cos 4 x)+\frac{1}{2}(1-\cos 6 x)=0 \Leftrightarrow$
$\Leftrightarrow 2 \cos 6 x \cos x-1-\cos 4 x+1-\cos 6 x=0, 2 \cos 6 x \cos x-$
$-(\cos 4 x+\cos 6 x)=0 \Leftrightarrow 2 \cos 6 x \cos x-2 \cos \frac{4 x+6 x}{2}... | x_{1}=\frac{\pi}{2}(2k+1),x_{2}=\frac{2\pin}{11},k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,934 |
8.033. $2(\cos 4 x-\sin x \cos 3 x)=\sin 4 x+\sin 2 x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.033. $2(\cos 4 x-\sin x \cos 3 x)=\sin 4 x+\sin 2 x$. | ## Solution.
## Given
$2 \cos 4 x-2 \sin x \cos 3 x-\sin 4 x-\sin 2 x=0, 2 \cos 4 x-\sin (x-3 x)-$ $-\sin (x+3 x)-\sin 4 x-\sin 2 x=0, 2 \cos 4 x+\sin 2 x-\sin 4 x-\sin 4 x$ $-\sin 2 x=0, 2 \cos 4 x-2 \sin 4 x=0 \Leftrightarrow \tan 4 x=1$,
from which $4 x=\frac{\pi}{4}+\pi k, \quad x=\frac{\pi}{16}+\frac{\pi k}{4}=... | \frac{\pi}{16}(4k+1),\quadk\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,935 |
8.034. $\sin x \cos x \cos 2 x \cos 8 x=\frac{1}{4} \sin 12 x$. | ## Solution.
From the condition
$$
\begin{aligned}
& 2(2 \sin x \cos x) \cos 2 x \cos 8 x=\sin 12 x \Leftrightarrow 2 \sin 2 x \cos 2 x \cos 8 x=\sin 12 x \Leftrightarrow \\
& \Leftrightarrow \sin 4 x \cos 8 x-\sin 12 x=0 \Leftrightarrow \frac{1}{2}(\sin (4 x-8 x)+\sin (4 x+8 x))-\sin 12 x=0, \\
& -\sin 4 x+\sin 12 x... | \frac{\pik}{8},k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 50,936 |
8.035. $3 \sin ^{2} 2 x+7 \cos 2 x-3=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.035. $3 \sin ^{2} 2 x+7 \cos 2 x-3=0$. | ## Solution.
We have
$$
\begin{aligned}
& 3\left(1-\cos ^{2} 2 x\right)+7 \cos 2 x-3=0 \Leftrightarrow 3 \cos ^{2} 2 x-7 \cos 2 x=0 \\
& \cos 2 x(3 \cos 2 x-7)=0
\end{aligned}
$$
From this:
1) $\cos 2 x=0, \quad 2 x=\frac{\pi}{2}+\pi k, \quad x_{1}=\frac{\pi}{4}+\frac{\pi k}{2}=\frac{\pi}{4}(2 k+1), k \in Z$;
2) $3... | \frac{\pi}{4}(2k+1),\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,937 |
8.036. $\sin 2 x \sin 6 x-\cos 2 x \cos 6 x=\sqrt{2} \sin 3 x \cos 8 x$. | Solution.
## From the condition
$(\cos 2 x \cos 6 x-\sin 2 x \sin 6 x)+\sqrt{2} \sin 3 x \cos 8 x=0 \Leftrightarrow$ $\Leftrightarrow \cos 8 x+\sqrt{2} \sin 3 x \cos 8 x=0, \quad \cos 8 x(1+\sqrt{2} \sin 3 x)=0$.
From this:
1) $\cos 8 x=0, 8 x=\frac{\pi}{2}+\pi n, x_{1}=\frac{\pi}{16}+\frac{\pi n}{8}=\frac{\pi}{16}... | x_{1}=\frac{\pi}{16}(2n+1),\quadx_{2}=(-1)^{k+1}\frac{\pi}{12}+\frac{\pik}{3},\quadn,k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 50,938 |
8.037. $\sin 3 x \cos 3 x=\sin 2 x$.
8.037. $\sin 3 x \cos 3 x=\sin 2 x$. | Solution.
We have
$2 \sin 3 x \cos 3 x-2 \sin 2 x=0, \Leftrightarrow \sin 6 x-2 \sin 2 x=0$,
$\sin 3(2 x)-2 \sin 2 x=0, \Leftrightarrow 3 \sin 2 x-4 \sin ^{3} 2 x-2 \sin 2 x=0$, $4 \sin ^{3} 2 x-\sin 2 x=0, \sin 2 x\left(4 \sin ^{2} 2 x-1\right)=0$.
## From this:
1) $\sin 2 x=0,2 x=\pi n, \quad x_{1}=\frac{\pi n}{... | x_{1}=\frac{\pin}{2};x_{2,3}=\frac{\pi}{12}(6k\1),n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,939 |
8.039. $3 \sin 2 x+2 \cos 2 x=3$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
8.039. $3 \sin 2 x+2 \cos 2 x=3$. | Solution.
We have
$6 \sin x \cos x+2\left(\cos ^{2} x-\sin ^{2} x\right)=3\left(\cos ^{2} x+\sin ^{2} x\right)$,
$5 \sin ^{2} x-6 \sin x \cos x+\cos ^{2} x=0, \Leftrightarrow 5 \operatorname{tg}^{2} x-6 \operatorname{tg} x+1=0$.
Solving this equation as a quadratic equation in terms of $\operatorname{tg} x$, we fin... | x_{1}=\operatorname{arcctg}5+\pik;x_{2}=\frac{\pi}{4}(4n+1),\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,940 |
8.040. $\operatorname{ctg}\left(\frac{3 \pi}{2}-x\right)-\operatorname{ctg}^{2} x+\frac{1+\cos 2 x}{\sin ^{2} x}=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0 .\end{array}\right.$
Rewrite the given equation as
$$
\begin{aligned}
& \operatorname{tg} x-\operatorname{ctg}^{2} x+\frac{1+2 \cos ^{2} x-1}{\sin ^{2} x}=0, \Leftrightarrow \operatorname{tg} x-\operatorname{ctg}^{2} x+2 \operato... | \frac{\pi}{4}(4k+3),\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,941 |
8.041. $\cos 9 x-\cos 7 x+\cos 3 x-\cos x=0$. | Solution.
Rewrite the equation as
$-2 \sin \frac{9 x+7 x}{2} \sin \frac{9 x-7 x}{2}-2 \sin \frac{3 x+x}{2} \sin \frac{3 x-x}{2}=0 \Leftrightarrow$
$\Leftrightarrow \sin 8 x \sin x+\sin 2 x \sin x=0 \Leftrightarrow \sin x(\sin 8 x+\sin 2 x)=0$.
From this:
1) $\sin x=0, \quad x_{1}=\pi n, \quad n \in Z$
2) $\sin 8 x... | x_{1}=\frac{\pik}{5};x_{2}=\frac{\pi}{6}(2+1),\quadk,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,942 |
8.042. $2\left(\operatorname{tg} \frac{t}{2}-1\right)=\cos t$.
8.042. $2\left(\tan \frac{t}{2}-1\right)=\cos t$. | ## Solution.
Domain of definition: $\cos \frac{t}{2} \neq 0$.
## We have
$$
\begin{aligned}
& 2 \cdot\left(\frac{\sin \frac{t}{2}}{\cos \frac{t}{2}}-1\right)-\left(\cos ^{2} \frac{t}{2}-\sin ^{2} \frac{t}{2}\right)=0 \Leftrightarrow \frac{2\left(\sin \frac{t}{2}-\cos \frac{t}{2}\right)}{\cos \frac{t}{2}}-\left(\cos ... | \frac{\pi}{2}(4n+1),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,943 |
8.043. $\sin 3z - \cos 3z = \sqrt{\frac{3}{2}}$. | Solution.
From the condition
$$
\begin{aligned}
& \sin 3 z \cdot \frac{\sqrt{2}}{2}-\cos 3 z \cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{3}}{2} \Leftrightarrow \sin 3 z \cos 45^{\circ}-\cos 3 z \sin 45^{\circ}=\frac{\sqrt{3}}{2} \Leftrightarrow \\
& \Leftrightarrow \sin \left(3 z-45^{\circ}\right)=\frac{\sqrt{3}}{2}
\end{al... | z_{1}=35+120k,z_{2}=55+120k,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,944 |
8.044. $\sqrt{3} \sin 2 x+\cos 5 x-\cos 9 x=0$.
8.044. $\sqrt{3} \sin 2 x+\cos 5 x-\cos 9 x=0$. | Solution.
Rewrite the given equation as
$$
\begin{aligned}
& \sqrt{3} \sin 2 x-2 \sin \frac{5 x+9 x}{2} \sin \frac{5 x-9 x}{2}=0 \Leftrightarrow \\
& \Leftrightarrow \sqrt{3} \sin 2 x+2 \sin 7 x \sin 2 x=0 \Leftrightarrow \sin 2 x(\sqrt{3}+2 \sin 7 x)=0
\end{aligned}
$$
From this:
1) $\sin 2 x=0, 2 x=\pi n, x_{1}=\... | x_{1}=\frac{\pin}{2};x_{2}=(-1)^{k+1}\frac{\pi}{21}+\frac{\pik}{7},n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,945 |
8.045. $2 \cos ^{2} x+5 \sin x-4=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8.045. $2 \cos ^{2} x+5 \sin x-4=0$. | ## Solution.
We have $2\left(1-\sin ^{2} x\right)+5 \sin x-4=0$ or $2 \sin ^{2} x-5 \sin x+2=0$. Solving this equation as a quadratic equation in terms of $\sin x$, we get $\sin x=2, \varnothing$, or $\sin x=\frac{1}{2}, x=(-1)^{k} \frac{\pi}{6}+\pi k, \quad k \in Z$.
Answer: $\quad x=(-1)^{k} \frac{\pi}{6}+\pi k, \q... | (-1)^{k}\frac{\pi}{6}+\pik,\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,946 |
8.047. $\sin ^{3} z \cos z-\sin z \cos ^{3} z=\frac{\sqrt{2}}{8}$. | ## Solution.
From the condition
$-\sin z \cos z\left(\cos ^{2} z-\sin ^{2} z\right)=\frac{\sqrt{2}}{8} \Leftrightarrow 2 \sin z \cos z\left(\cos ^{2} z-\sin ^{2} z\right)=-\frac{\sqrt{2}}{4} \Leftrightarrow$ $\Leftrightarrow \sin 2 z \cos 2 z=-\frac{\sqrt{2}}{4} \Leftrightarrow 2 \sin 2 z \cos 2 z=-\frac{\sqrt{2}}{2}... | (-1)^{k+1}\frac{\pi}{16}+\frac{\pik}{4},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,948 |
8.048. $\sin \left(\frac{\pi}{4}+5 x\right) \cos \left(\frac{\pi}{4}+2 x\right)=\sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-6 x\right)$ | ## Solution.
Rewrite the given equation as
$$
\begin{aligned}
& \frac{1}{2}\left(\sin \left(\frac{\pi}{4}+5 x-\frac{\pi}{4}-2 x\right)+\sin \left(\frac{\pi}{4}+5 x+\frac{\pi}{4}+2 x\right)\right)= \\
& =\frac{1}{2}\left(\cos \left(\frac{\pi}{4}+x-\frac{\pi}{4}+6 x\right)-\cos \left(\frac{\pi}{4}+x+\frac{\pi}{4}-6 x\r... | \frac{\pin}{4},n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,949 |
8.049. $\cos 3 x=2 \sin \left(\frac{3 \pi}{2}+x\right)$
Translate the above text into English, preserving the original text's line breaks and format, and output the translation result directly.
8.049. $\cos 3 x=2 \sin \left(\frac{3 \pi}{2}+x\right)$ | ## Solution.
Let's rewrite the given equation as
$$
\begin{aligned}
& 4 \cos ^{3} x-3 \cos x=-2 \cos x, \Leftrightarrow 4 \cos ^{3} x-\cos x=0 \Leftrightarrow \\
& \Leftrightarrow \cos x\left(4 \cos ^{2} x-1\right)=0
\end{aligned}
$$
## From this
1) $\cos x=0, \quad x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k... | x_{1}=\frac{\pi}{2}(2k+1),\quadx_{2}=\\frac{\pi}{3}+\pin,\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,950 |
8.051. $1+\sin 2 x=(\cos 3 x+\sin 3 x)^{2}$. | ## Solution.
We have
$$
\begin{aligned}
& 1+\sin 2 x=\cos ^{2} 3 x+2 \sin 3 x \cos 3 x+\sin ^{2} 3 x, \quad 1+\sin 2 x=1+\sin 6 x \\
& \sin 6 x-\sin 2 x=0 \Leftrightarrow 2 \cos \frac{6 x+2 x}{2} \sin \frac{6 x-2 x}{2}=0, \quad \cos 4 x \sin 2 x=0
\end{aligned}
$$
from which:
1) $\cos 4 x=0, \quad 4 x=\frac{\pi}{2}... | x_{1}=\frac{\pi}{8}(2k+1),x_{2}=\frac{\pin}{2},\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,951 |
8.052. $\sin 3 x=2 \cos \left(\frac{\pi}{2}-x\right)$ | ## Solution.
## Rewrite the equation as
$3 \sin x-4 \sin ^{3} x=2 \sin x, 4 \sin ^{3} x-\sin x=0, \quad \sin x\left(4 \sin ^{2} x-1\right)=0$, from which:
1) $\sin x=0, x_{1}=\pi n, \quad n \in Z$
2) $4 \sin ^{2} x-1=0, \quad \sin x= \pm \frac{1}{2}, \quad x_{2,3}= \pm \frac{\pi}{6}+\pi k, \quad k \in Z$.
Answer: $... | x_{1}=\pin;x_{2,3}=\\frac{\pi}{6}+\pik,\quadn,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,952 |
8.053. $\cos 4 x+2 \sin ^{2} x=0$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
8.053. $\cos 4 x+2 \sin ^{2} x=0$. | Solution.
## Rewrite the equation as
$$
\begin{aligned}
& 2 \cos ^{2} 2 x-1+1-\cos 2 x=0, \quad 2 \cos ^{2} 2 x-\cos 2 x=0 \\
& \cos 2 x(2 \cos 2 x-1)=0
\end{aligned}
$$
from which:
1) $\cos 2 x=0, \quad 2 x=\frac{\pi}{2}+\pi k, \quad x_{1}=\frac{\pi}{4}+\frac{\pi k}{2}=\frac{\pi}{4}(2 k+1), k \in Z$;
2) $2 \cos 2 ... | x_{1}=\frac{\pi}{4}(2k+1),\quadx_{2}=\\frac{\pi}{6}+\pin,\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,953 |
8.054. $\sin x+\sin 7 x-\cos 5 x+\cos (3 x-2 \pi)=0$. | ## Solution.
From the condition
$$
(\sin x+\sin 7 x)-(\cos 5 x-\cos 3 x)=0 \Leftrightarrow 2 \sin \frac{x+7 x}{2} \cos \frac{x-7 x}{2}+
$$
$+2 \sin \frac{5 x+3 x}{2} \sin \frac{5 x-3 x}{2}=0 \Leftrightarrow \sin 4 x \cos x+\sin 4 x \sin x=0$,
$\sin 4 x(\cos 3 x+\sin x)=0$,
from which:
1) $\sin 4 x=0,4 x=\pi k, x_... | x_{1}=\frac{\pik}{4},x_{2}=\frac{\pi}{8}(4n+3),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,954 |
8.055. $\cos ^{4} 2 x+6 \cos ^{2} 2 x=\frac{25}{16}$. | ## Solution.
We have $16 \cos ^{4} 2 x+96 \cos ^{2} 2 x-25=0$. Solving this equation as a biquadratic equation in terms of $\cos 2 x$, we get $\cos 2 x= \pm \frac{1}{2}, 2 x= \pm \frac{\pi}{3}+\pi k$, $x= \pm \frac{\pi}{6}+\frac{\pi k}{2}$, where $k \in Z$.
Answer: $\quad x= \pm \frac{\pi}{6}+\frac{\pi k}{2}, k \in Z... | \\frac{\pi}{6}+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,955 |
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