8.200. $\frac{3(\cos 2 x+\operatorname{ctg} 2 x)}{\operatorname{ctg} 2 x-\cos 2 x}-2(\sin 2 x+1)=0$.
## Solution.

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

From the condition we have:

$$
\begin{aligned}
& \frac{3\left(\cos 2 x+\frac{\cos 2 x}{\sin 2 x}\right)}{\frac{\cos 2 x}{\sin 2 x}-\cos 2 x}-2(\sin 2 x+1)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{3 \cos 2 x(\sin 2 x+1)}{\cos 2 x(1-\sin 2 x)}-2(\sin 2 x+1)=0 \Leftrightarrow(\sin 2 x+1)\left(\frac{3}{1-\sin 2 x}-2\right)=0
\end{aligned}
$$

## From here

1) $\sin 2 x+1=0, \sin 2 x=-1,2 x=-\frac{\pi}{2}+2 \pi k, x_{1}=-\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k-1)$, $k \in Z ;$
2) $\frac{3}{1-\sin 2 x}-2=0, \sin 2 x=-\frac{1}{2}, 2 x=(-1)^{n+1} \frac{\pi}{6}+\pi n$, $x_{2}=(-1)^{n+1} \frac{\pi}{12}+\frac{\pi n}{2} n \in Z ; x_{1}$ does not fit the domain of definition.

Answer: $x=(-1)^{k+1} \frac{\pi}{12}+\frac{\pi k}{2}, k \in Z$.