8.300. $\operatorname{tg} 5 x-2 \operatorname{tg} 3 x=\operatorname{tg}^{2} 3 x \operatorname{tg} 5 x$.
Solution.

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

From the condition we have

$$
\begin{aligned}
& \operatorname{tg} 5 x-\operatorname{tg} 3 x=\operatorname{tg}^{2} 3 x \operatorname{tg} 5 x+\operatorname{tg} 3 x \Leftrightarrow \\
& \Leftrightarrow \operatorname{tg} 5 x-\operatorname{tg} 3 x=\operatorname{tg} 3 x(\operatorname{tg} 3 x \operatorname{tg} 5 x+1) \Leftrightarrow
\end{aligned}
$$

$$
\begin{aligned}
& \Leftrightarrow \frac{\sin (5 x-3 x)}{\cos 5 x \cos 3 x}-\frac{\sin 3 x}{\cos 3 x}\left(\frac{\sin 3 x \sin 5 x+\cos 3 x \cos 5 x}{\cos 3 x \cos 5 x}\right)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\sin 2 x}{\cos 5 x \cos 3 x}-\frac{\sin 3 x \cos 2 x}{\cos ^{2} 3 x \cos 5 x}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\sin 2 x \cos 3 x-\cos 2 x \sin 3 x}{\cos ^{2} 3 x \cos 5 x}=0 \Leftrightarrow \frac{-\sin x}{\cos ^{2} 3 x \cos 5 x}=0 \Rightarrow \sin x=0 \\
& x=\pi k, k \in Z
\end{aligned}
$$

Answer: $x=\pi k, k \in Z$