8.282. $5 \sin ^{4} 2 z-4 \sin ^{2} 2 z \cos ^{2} 2 z-\cos ^{4} 2 z+4 \cos 4 z=0$.
Solution.

Rewrite the equation as

$5\left(\sin ^{2} 2 z\right)^{2}-\sin ^{2} 4 z-\left(\cos ^{2} 2 z\right)^{2}+4 \cos 4 z=0 \Leftrightarrow$

$\Leftrightarrow 5 \cdot\left(\frac{1}{2}(1-\cos 4 z)\right)^{2}-\left(1-\cos ^{2} 4 z\right)-\left(\frac{1}{2}(1-\cos 4 z)\right)^{2}+4 \cos 4 z=0 \Leftrightarrow$

$\Leftrightarrow 2 \cos ^{2} 4 z+\cos 4 z=0, \cos 4 z(2 \cos 4 z+1)=0$.

From this,

1) $\cos 4 z=0,4 z=\frac{\pi}{2}+\pi k, z_{1}=\frac{\pi}{8}+\frac{\pi k}{4}=\frac{\pi}{8}(2 k+1), k \in Z$;
2) $2 \cos 4 z+1=0, \cos 4 z=-\frac{1}{2}, 4 z= \pm \frac{2}{3} \pi+2 \pi n$,

$z_{2}= \pm \frac{\pi}{6}+\frac{\pi n}{2}=\frac{\pi}{6}(3 n \pm 1), n \in Z$

Answer: $z_{1}=\frac{\pi}{8}(2 k+1) ; z_{2}=\frac{\pi}{6}(3 n \pm 1)$, where $k$ and $n \in Z$.