### 8.136. $\sin 2 z-4 \cos 2 z=4$.
## Solution.

Let's rewrite the given equation as

$$
2 \sin z \cos z-4\left(\cos ^{2} z-\sin ^{2} z\right)-4\left(\cos ^{2} z+\sin ^{2} z\right)=0
$$

$$
2 \sin z \cos z-8 \cos ^{2} z=0, \quad 2 \cos z(\sin z-4 \cos z)=0
$$

## From this:

1) $\cos z=0, \quad z_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$
2) $\sin z-4 \cos z=0 \Leftrightarrow \operatorname{tg} z=4, \quad z_{2}=\operatorname{arctg} 4+\pi n, \quad n \in Z$.

Answer: $\quad z_{1}=\frac{\pi}{2}(2 k+1), z_{2}=\operatorname{arctg} 4+\pi n, k, n \in Z$.