8.304. $\operatorname{tg} 2 x-\operatorname{ctg} 3 x+\operatorname{ctg} 5 x=0$.

8.304. $\tan 2x - \cot 3x + \cot 5x = 0$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\cos 2 x \neq 0, \\ \sin 3 x \neq 0, \\ \sin 5 x \neq 0 .\end{array}\right.$

Rewrite the equation as

$$
\begin{aligned}
& \left(\frac{\sin 2 x}{\cos 2 x}-\frac{\cos 3 x}{\sin 3 x}\right)+\frac{\cos 5 x}{\sin 5 x}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\sin 2 x \sin 3 x-\cos 2 x \cos 3 x}{\cos 2 x \sin 3 x}+\frac{\cos 5 x}{\sin 5 x}=0 \Leftrightarrow \\
& \Leftrightarrow-\frac{\cos 5 x}{\cos 2 x \sin 3 x}+\frac{\cos 5 x}{\sin 5 x}=0 \Leftrightarrow-\cos 5 x(2 \sin 5 x-2 \cos 2 x \sin 3 x)=0
\end{aligned}
$$

From this

1) $\cos 5 x=0,5 x=\frac{\pi}{2}+\pi k, x_{1}=\frac{\pi}{10}+\frac{\pi k}{5}=\frac{\pi}{10}(2 k+1), k \in Z$;
2) $2 \sin 5 x-2 \cos 2 x \sin 3 x=0 \Leftrightarrow 2 \sin 5 x-\sin x-\sin 5 x=0 \Leftrightarrow$ $\Leftrightarrow \sin 5 x-\sin x=0 \Leftrightarrow 2 \sin 2 x \cos 3 x=0$.

Then either $\sin 2 x=0,2 x=\pi n, x_{2}=\frac{\pi n}{2}, n \in Z$, or $\cos 3 x=0,3 x=\frac{\pi}{2}+\pi l$, $x_{3}=\frac{\pi}{6}+\frac{\pi l}{3}, l \in Z ; x_{2}$ is included in $x_{1}$ or in $x_{3}$.

Answer: $x_{1}=\frac{\pi}{10}(2 k+1) ; x_{2}=\frac{\pi}{6}(2 l+1)$, where $k$ and $l \in Z$.