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

$\frac{\sin 2 z}{\cos 5 z \cos 3 z}-\frac{2 \sin 2 z}{\cos 2 z}=0 \Leftrightarrow \sin 2 z(\cos 2 z-2 \cos 5 z \cos 3 z)=0 \Leftrightarrow$

$\Leftrightarrow \sin 2 z \cdot(\cos 2 z-\cos 2 z-\cos 8 z)=0,-\sin 2 z \cos 8 z=0$.

From this

1) $\sin 2 z=0,2 z=\pi k, z_{1}=\frac{\pi k}{2}, k \in Z$
2) $\cos 8 z=0,8 z=\frac{\pi}{2}+\pi n, z_{2}=\frac{\pi}{16}+\frac{\pi n}{8}=\frac{\pi}{16}(2 n+1), n \in Z$.

Considering the domain of definition, we get $z_{1}=\pi k, k \in Z$.

Answer: $z_{1}=\pi k ; z_{2}=\frac{\pi}{16}(2 n+1), k$ and $n \in Z$.