8.148. $\operatorname{tg} 2 x \sin 2 x-3 \sqrt{3} \operatorname{ctg} 2 x \cos 2 x=0$.
## Solution.

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

## We have

$$
\begin{aligned}
& \frac{\sin 2 x \sin 2 x}{\cos 2 x}-\frac{3 \sqrt{3} \cos 2 x \cos 2 x}{\sin 2 x}=0 \Rightarrow \sin ^{3} 2 x-3 \sqrt{3} \cos ^{3} 2 x=0 \\
& \operatorname{tg}^{3} 2 x=3 \sqrt{3}, \operatorname{tg} 2 x=\sqrt{3}, 2 x=\frac{\pi}{3}+\pi n, x=\frac{\pi}{6}+\frac{\pi n}{2}=\frac{\pi}{6}(3 n+1), n \in Z
\end{aligned}
$$

Answer: $\quad x=\frac{\pi}{6}(3 n+1), n \in Z$.