8.054. $\sin x+\sin 7 x-\cos 5 x+\cos (3 x-2 \pi)=0$.
## Solution.

From the condition

$$
(\sin x+\sin 7 x)-(\cos 5 x-\cos 3 x)=0 \Leftrightarrow 2 \sin \frac{x+7 x}{2} \cos \frac{x-7 x}{2}+
$$

$+2 \sin \frac{5 x+3 x}{2} \sin \frac{5 x-3 x}{2}=0 \Leftrightarrow \sin 4 x \cos x+\sin 4 x \sin x=0$,

$\sin 4 x(\cos 3 x+\sin x)=0$,

from which:

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

The obtained equation is equivalent to two equations:

a) $\cos \left(x+\frac{\pi}{4}\right)=0, \quad x+\frac{\pi}{4}=\frac{\pi}{2}+\pi n, \quad x_{2}=\frac{\pi}{4}+\pi n, n \in Z$;

b) $\cos \left(2 x-\frac{\pi}{4}\right)=0,2 x-\frac{\pi}{4}=\frac{\pi}{2}+\pi m, \quad x_{3}=\frac{\pi}{8}(4 m+3), m \in Z$.

## Solutions $x_{2}$ are included in $x_{1}$.

Answer: $\quad x_{1}=\frac{\pi k}{4}, x_{2}=\frac{\pi}{8}(4 n+3), k, n \in Z$.