8.070. $3 \sin 5 z-2 \cos 5 z=3$.

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8.070. $3 \sin 5 z-2 \cos 5 z=3$.
Solution.

From the condition

$$
3 \sin 2\left(\frac{5}{2} z\right)-2 \cos 2\left(\frac{5}{2} z\right)-3\left(\cos ^{2} \frac{5}{2} z+\sin ^{2} \frac{5}{2} z\right)=0 \Leftrightarrow
$$

$\Leftrightarrow 6 \sin \frac{5}{2} z \cos \frac{5}{2} z-2\left(\cos ^{2} \frac{5}{2} z-\sin ^{2} \frac{5}{2} z\right)-3\left(\cos ^{2} \frac{5}{2} z+\sin ^{2} \frac{5}{2} z\right)=0 \Leftrightarrow$ $\Leftrightarrow \sin ^{2} \frac{5}{2} z-6 \sin \frac{5}{2} z \cos \frac{5}{2} z+5 \cos ^{2} \frac{5}{2} z=0$.

Dividing this equation by $\cos ^{2} \frac{5}{2} z \neq 0$, we get

$\operatorname{tg}^{2} \frac{5}{2} z-6 \operatorname{tg} \frac{5}{2} z+5=0$

Solving the equation as a quadratic equation in $\operatorname{tg} \frac{5}{2} z$, we find

$\operatorname{tg} \frac{5}{2} z=1, \frac{5}{2} z=\frac{\pi}{4}+\pi k, z_{1}=\frac{\pi}{10}+\frac{2}{5} \pi k, k \in Z$,

or

$\operatorname{tg} \frac{5}{2} z=5, \frac{5}{2} z=\operatorname{arctg} 5+\pi n, z_{2}=\frac{2}{5} \operatorname{arctg} 5+\frac{2}{5} \pi n, n \in Z$.

Answer: $\quad z_{1}=\frac{\pi}{10}+\frac{2}{5} \pi k ; z_{2}=\frac{2}{5} \operatorname{arctg} 5+\frac{2}{5} \pi n, \quad k, n \in Z$.