8.115. $\frac{\sin z}{1+\cos z}=2-\operatorname{ctg} z$.

8.115. $\frac{\sin z}{1+\cos z}=2-\cot z$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\sin z \neq 0, \\ \cos z \neq-1 .\end{array}\right.$

From the condition

$\frac{\sin z}{1+\cos z}-2+\frac{\cos z}{\sin z}=0 \Rightarrow \sin ^{2} z-2 \sin z(1+\cos z)+\cos z(1+\cos z)=0$, $\sin ^{2} z-2 \sin z-2 \sin z \cos z+\cos z+\cos ^{2} z=0,(1-2 \sin z)+$ $+(\cos z-2 \sin z \cos z)=0, \quad(1-2 \sin z)+\cos z(1-2 \sin z)=0$, $(1-2 \sin z)(1+\cos z)=0$.

From this

$1-2 \sin z=0, \quad \sin z=\frac{1}{2}, \quad z=(-1)^{k} \frac{\pi}{6}+\pi k, \quad k \in Z ; 1+\cos z \neq 0$.

Answer: $\quad z=(-1)^{k} \frac{\pi}{6}+\pi k, \quad k \in Z$.