$5.35 \operatorname{tg} 2 x-\operatorname{ctg} 3 x+\operatorname{ctg} 5 x=0$.

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

$5.35 \tan 2x - \cot 3x + \cot 5x = 0$.
### 5.35 We have

$$
\begin{aligned}
& \frac{\sin 2 x}{\cos 2 x} - \frac{\cos 3 x}{\sin 3 x} + \frac{\cos 5 x}{\sin 5 x} = 0 \\
& \sin 3 x \neq 0, \sin 5 x \neq 0, \cos 2 x \neq 0 \\
& \frac{\sin 2 x \sin 3 x - \cos 2 x \cos 3 x}{\cos 2 x \sin 3 x} + \frac{\cos 5 x}{\sin 5 x} = 0
\end{aligned}
$$

$-\frac{\cos 5 x}{\cos 2 x \sin 3 x} + \frac{\cos 5 x}{\sin 5 x} = 0$.

Further, we get:

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

$x_{2} = \frac{\pi k}{2}, k \neq 2 l$, since in this case $\sin 3 x = \sin 5 x = 0$;

$x_{3} = \frac{\pi}{6}(2 k + 1)$ includes the solution $x_{2}$.

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