8.320. $\sin 2z - \sin 6z + 2 = 0$.
Solution.

Write the equation as $\sin 2 z-\sin 3(2 z)+2=0$. Using the formula $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we have

$$
\begin{aligned}
& \sin 2 z-3 \sin 2 z+4 \sin ^{3} 2 z+2=0,4 \sin ^{3} 2 z-2 \sin 2 z+2=0 \\
& 2 \sin ^{3} 2 z-\sin 2 z+1=0,2 \sin ^{3} 2 z+2-\sin 2 z-1=0
\end{aligned}
$$

$2(\sin 2 z+1)\left(\sin ^{2} 2 z-\sin 2 z+1\right)-(\sin 2 z+1)=0$, $(\sin 2 z+1)\left(2 \sin ^{2} 2 z-2 \sin 2 z+1\right)=0$.

## Hence

1) $\sin 2 z+1=0, \sin 2 z=-1,2 z=-\frac{\pi}{2}+2 \pi k, z=-\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k-1), k \in Z$,
2) $2 \sin ^{2} 2 z-2 \sin 2 z+1=0(D<0)$, $\varnothing$.

Answer: $z=\frac{\pi}{4}(4 k-1), k \in Z$.