8.069. $2 \sin ^{3} x-\cos 2 x-\sin x=0$.

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8.069. $2 \sin ^{3} x-\cos 2 x-\sin x=0$.
## Solution.

We have

$$
\begin{aligned}
& 2 \sin ^{3} x-1+2 \sin ^{2} x-\sin x=0, \quad 2 \sin ^{3} x+2 \sin ^{2} x-\sin x-1=0, \\
& 2 \sin ^{2} x(\sin x+1)-(\sin x+1)=0, \quad(\sin x+1)\left(2 \sin ^{2} x-1\right)=0 .
\end{aligned}
$$

From this:

1) $\sin x+1=0, \quad \sin x=-1, \quad x_{1}=-\frac{\pi}{2}+2 \pi k=\frac{\pi}{2}(4 k-1) \quad k \in \mathbb{Z}$;
2) $2 \sin ^{2} x-1=0, \sin x= \pm \frac{\sqrt{2}}{2} ; \quad x_{2}=-\frac{\pi}{4}+\frac{\pi n}{2}=\frac{\pi}{4}(2 n+1), \quad n \in \mathbb{Z}$.

Answer: $\quad x_{1}=\frac{\pi}{2}(4 k-1), x_{2}=\frac{\pi}{4}(2 n+1) \quad k, n \in \mathbb{Z}$.