8.071. $4 \sin 3 z+\frac{1}{3} \cos 3 z=3$.
## Solution.

## Rewrite the equation as

$$
\begin{aligned}
& 24 \sin \frac{3 z}{2} \cos \frac{3 z}{2}+\cos ^{2} \frac{3 z}{2}-\sin ^{2} \frac{3 z}{2}-9 \cos ^{2} \frac{3 z}{2}-9 \sin ^{2} \frac{3 z}{2}=0 \Leftrightarrow \\
& \Leftrightarrow 10 \sin ^{2} \frac{3 z}{2}-24 \sin \frac{3 z}{2} \cos \frac{3 z}{2}+8 \cos ^{2} \frac{3 z}{2}=0
\end{aligned}
$$

Dividing this equation by \(2 \cos ^{2} \frac{3 z}{2} \neq 0\), we get

$$
5 \operatorname{tg}^{2} \frac{3 z}{2}-12 \operatorname{tg} \frac{3 z}{2}+4=0
$$

Solving the equation as a quadratic in \(\operatorname{tg} \frac{3 z}{2}\), we find \(\operatorname{tg} \frac{3 z}{2}=\frac{2}{5}\), \(\frac{3 z}{2}=\operatorname{arctg} \frac{2}{5}+\pi n, z_{1}=\frac{2}{3} \operatorname{arctg} \frac{2}{5}+\frac{2}{3} \pi n, n \in Z ; \operatorname{tg} \frac{3 z}{2}=2, \frac{3}{2} x=\operatorname{arctg} 2+\pi k, z_{2}=\frac{2}{3} \operatorname{arctg} 2+\frac{2}{3} \pi k, k \in Z\).

Answer: \(\quad z_{1}=\frac{2}{3} \operatorname{arctg} \frac{2}{5}+\frac{2}{3} \pi n ; z_{2}=\frac{2}{3} \operatorname{arctg} 2+\frac{2}{3} \pi k, \quad k, n \in Z\).