8.015. $\sin ^{2} 2 z+\sin ^{2} 3 z+\sin ^{2} 4 z+\sin ^{2} 5 z=2$.
## Solution.

Rewrite the equation as

$$
\begin{aligned}
& \frac{1}{2}(1-\cos 4 z)+\frac{1}{2}(1-\cos 6 z)+\frac{1}{2}(1-\cos 8 z)+\frac{1}{2}(1-\cos 10 z)=2 \\
& (\cos 4 z+\cos 6 z)+(\cos 8 z+\cos 10 z)=0, \Leftrightarrow 2 \cos 5 z \cos z+2 \cos 9 z \cos z=0
\end{aligned}
$$

$2 \cos z(\cos 5 z+\cos 9 z)=0$.

Then:

1) $\cos z=0, z_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$,

or
2) $\cos 5 z+\cos 9 z=0,2 \cos 7 z \cos 2 z=0, \cos 7 z=0,7 z=\frac{\pi}{2}+\pi n$,

$$
\begin{aligned}
& z_{2}=\frac{\pi}{14}+\frac{\pi n}{7}=\frac{\pi}{14}(2 n+1), \quad n \in Z ; \cos 2 z=0,2 z=\frac{\pi}{2}+\pi m \\
& z_{3}=\frac{\pi}{4}+\frac{\pi m}{2}=\frac{\pi}{4}(2 m+1), \quad m \in Z ; z_{1} \text { is included in } z_{2}
\end{aligned}
$$

Answer: $\quad z_{1}=\frac{\pi}{14}(2 n+1), z_{2}=\frac{\pi}{4}(2 m+1) \quad n, m \in Z$.