8.266. $\cos ^{3} x+\frac{1}{2} \sin 2 x-\cos x \sin ^{3} x+4 \sin x+4=0$.
## Solution.

Let's rewrite the equation as

$\cos ^{3} x+\sin x \cos x-\cos x \sin x \sin ^{2} x+4(\sin x+1)=0 \Leftrightarrow$

$\Leftrightarrow \cos ^{3} x+\sin x \cos x\left(1-\sin ^{2} x\right)+4(\sin x+1)=0 \Leftrightarrow$

$\Leftrightarrow \cos ^{3} x+\sin x \cos ^{3} x+4(\sin x+1)=0 \Leftrightarrow$

$\Leftrightarrow \cos ^{3} x(1+\sin x)+4(\sin x+1)=0 \Leftrightarrow$

$\Leftrightarrow(\sin x+1)\left(\cos ^{3} x+4\right)=0$.

From this,

$\sin x+1=0, \sin x=-1, x=-\frac{\pi}{2}+2 \pi k=\frac{\pi}{2}(4 k-1), k \in Z, \cos ^{3} x+4 \neq 0$.

Answer: $x=\frac{\pi}{2}(4 k-1), k \in Z$.