8.197. $\frac{\cot 2z}{\cot z}+\frac{\cot z}{\cot 2z}+2=0$.
## Solution.

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

From the condition we have:

$$
\begin{aligned}
& \left(\frac{\operatorname{ctg} 2 z}{\operatorname{ctg} z}\right)^{2}+2 \cdot \frac{\operatorname{ctg} 2 z}{\operatorname{ctg} z}+1=0 \Leftrightarrow\left(\frac{\operatorname{ctg} 2 z}{\operatorname{ctg} z}+1\right)^{2}=0 \Leftrightarrow \frac{\operatorname{ctg} 2 z}{\operatorname{ctg} z}=-1, \\
& \operatorname{ctg} 2 z=-\operatorname{ctg} z, \operatorname{ctg} 2 z+\operatorname{ctg} z=0
\end{aligned}
$$

Since $\operatorname{ctg} \alpha+\operatorname{ctg} \beta=\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}$, we get $\frac{\sin 3 z}{\sin 2 z \sin z}=0$.

Using the formula $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we have

$$
\frac{3 \sin z-4 \sin ^{3} z}{\sin 2 z \sin z}=\frac{\sin z\left(3-4 \sin ^{2} z\right)}{\sin 2 z \sin z}=\frac{3-4 \sin ^{2} z}{\sin 2 z}
$$

and our equation becomes $3-4 \sin ^{2} z=0$, from which $\sin z= \pm \frac{\sqrt{3}}{2}$, $z= \pm \frac{\pi}{3}+\pi k=\frac{\pi}{3}(3 k \pm 1), k \in Z$

Answer: $z=\frac{\pi}{3}(3 k \pm 1), k \in Z$