8.246. $3 \sin ^{2} z \cos ^{2}\left(\frac{\pi}{2}+z\right)-\frac{1}{2} \sin ^{2} 2 z-5 \cos ^{4} z+2 \cos 2 z=0$.
## Solution.

From the condition we have:

$6\left(\sin ^{2} z\right)^{2}-\left(1-\cos ^{2} 2 z\right)-10\left(\cos ^{2} z\right)^{2}+4 \cos 2 z=0$.

Since $\sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2}$ and $\cos ^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{2}$, we get

$6 \cdot \frac{1}{4}(1-\cos 2 z)^{2}-1+\cos ^{2} 2 z-10 \cdot \frac{1}{4}(1+\cos 2 z)^{2}+4 \cos 2 z=0 \Leftrightarrow$

$\Leftrightarrow 3(1-\cos 2 z)^{2}-2+2 \cos ^{2} 2 z-5(1+\cos 2 z)^{2}+8 \cos 2 z=0 \Leftrightarrow$ $\Leftrightarrow \cos 2 z=-\frac{1}{2}, 2 z= \pm \frac{2}{3} \pi+2 \pi k, z= \pm \frac{\pi}{3}+2 \pi k=\frac{\pi}{3}(3 k \pm 1), k \in Z$.

Answer: $z=\frac{\pi}{3}(3 k \pm 1), k \in Z$.