8.279. $2 \cos z \sin ^{3}\left(\frac{3 \pi}{2}-z\right)-5 \sin ^{2} z \cos ^{2} z+\sin z \cos ^{3}\left(\frac{3 \pi}{2}+z\right)=\cos 2 z$.
## Solution.

From the condition we have

$-2 \cos z \cos ^{3} z-5 \sin ^{2} z \cos ^{2} z+\sin z \sin ^{3} z-\cos 2 z=0 \Leftrightarrow$

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

$\Leftrightarrow-2\left(1-\sin ^{2} z\right)^{2}-5 \sin ^{2} z\left(1-\sin ^{2} z\right)+\sin ^{4} z-\left(1-\sin ^{2} z\right)+\sin ^{2} z=0 \Leftrightarrow$ $\Leftrightarrow-2+4 \sin ^{2} z-2 \sin ^{4} z-5 \sin ^{2} z+5 \sin ^{4} z+\sin ^{4} z-1+\sin ^{2} z+\sin ^{2} z=0 \Leftrightarrow$ $\Leftrightarrow 4 \sin ^{4} z+\sin ^{2} z-3=0$.

Solving this equation as a quadratic in $\sin ^{2} z$, we get $\sin ^{2} z=-1, \varnothing ; \sin ^{2} z=\frac{3}{4} \Rightarrow \sin z= \pm \frac{\sqrt{3}}{2}$,

$z= \pm \frac{\pi}{3}+\pi n=\frac{\pi}{3}(3 n \pm 1), n \in Z$.

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