8.033. $2(\cos 4 x-\sin x \cos 3 x)=\sin 4 x+\sin 2 x$.

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8.033. $2(\cos 4 x-\sin x \cos 3 x)=\sin 4 x+\sin 2 x$.
## Solution.

## Given

$2 \cos 4 x-2 \sin x \cos 3 x-\sin 4 x-\sin 2 x=0, 2 \cos 4 x-\sin (x-3 x)-$ $-\sin (x+3 x)-\sin 4 x-\sin 2 x=0, 2 \cos 4 x+\sin 2 x-\sin 4 x-\sin 4 x$ $-\sin 2 x=0, 2 \cos 4 x-2 \sin 4 x=0 \Leftrightarrow \tan 4 x=1$,

from which $4 x=\frac{\pi}{4}+\pi k, \quad x=\frac{\pi}{16}+\frac{\pi k}{4}=\frac{\pi}{16}(4 k+1) \quad k \in \mathbb{Z}$.

Answer: $\quad x=\frac{\pi}{16}(4 k+1), \quad k \in \mathbb{Z}$.