8.180. $\frac{\cos ^{2} z(1+\operatorname{ctg} z)-3}{\sin z-\cos z}=3 \cos z$.
Solution.

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

Rewrite the equation as

$$
\begin{aligned}
& \cos ^{2} z\left(1+\frac{\cos z}{\sin z}\right)-3=3 \cos z(\sin z-\cos z) \Leftrightarrow \frac{\cos ^{2} z(\sin z+\cos z)}{\sin z}- \\
& -3-3 \cos z(\sin z-\cos z)=0 \Leftrightarrow \cos ^{2} z \sin z+\cos ^{3} z-3 \sin z- \\
& -3 \cos z \sin ^{2} z+3 \cos ^{2} z \sin z=0 \Leftrightarrow 4 \cos ^{2} z \sin z+\cos ^{3} z- \\
& -3 \sin z\left(\cos ^{2} z+\sin ^{2} z\right)-3 \cos z \sin ^{2} z=0 \Leftrightarrow \cos ^{2} z \sin z+\cos ^{3} z- \\
& -3 \sin ^{3} z-3 \cos z \sin ^{2} z=0 \Leftrightarrow \cos ^{2} z(\sin z+\cos z)- \\
& -3 \sin ^{2} z(\cos z+\sin z)=0 \Leftrightarrow(\sin z+\cos z)\left(\cos ^{2} z-3 \sin ^{2} z\right)=0
\end{aligned}
$$

From this, either $\sin z+\cos z=0$, or $\cos ^{2} z-3 \sin ^{2} z=0$. Dividing the first equation by $\sin z \neq 0$, and the second by $\sin ^{2} z \neq 0$, we get

$$
\operatorname{ctg} z=-1, z_{1}=-\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n-1), n \in Z
$$

or

$$
\operatorname{ctg}^{2} z=3, \operatorname{ctg} z= \pm \sqrt{3}, z_{2}= \pm \frac{\pi}{6}+\pi k, k \in Z
$$

Answer: $z_{1}=\frac{\pi}{4}(4 n-1) ; z_{2}= \pm \frac{\pi}{6}+\pi k$, where $n$ and $k \in Z$.