8.053. $\cos 4 x+2 \sin ^{2} x=0$.

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8.053. $\cos 4 x+2 \sin ^{2} x=0$.
Solution.

## Rewrite the equation as

$$
\begin{aligned}
& 2 \cos ^{2} 2 x-1+1-\cos 2 x=0, \quad 2 \cos ^{2} 2 x-\cos 2 x=0 \\
& \cos 2 x(2 \cos 2 x-1)=0
\end{aligned}
$$

from which:

1) $\cos 2 x=0, \quad 2 x=\frac{\pi}{2}+\pi k, \quad x_{1}=\frac{\pi}{4}+\frac{\pi k}{2}=\frac{\pi}{4}(2 k+1), k \in Z$;
2) $2 \cos 2 x-1=0, \cos 2 x=\frac{1}{2}, 2 x= \pm \frac{\pi}{3}+2 \pi n, x_{2}= \pm \frac{\pi}{6}+\pi n, n \in Z$.

Answer: $\quad x_{1}=\frac{\pi}{4}(2 k+1) x_{2}= \pm \frac{\pi}{6}+\pi n, \quad k, n \in Z$.