8.082. $\cos 7 x+\sin 8 x=\cos 3 x-\sin 2 x$.

8.082. $\cos 7 x+\sin 8 x=\cos 3 x-\sin 2 x$.
Solution.

We have

$(\cos 7 x-\cos 3 x)+(\sin 8 x+\sin 2 x)=0 \Leftrightarrow-2 \sin \frac{7 x+3 x}{2} \sin \frac{7 x-3 x}{2}+$

$+2 \sin \frac{8 x+2 x}{2} \cos \frac{8 x-2 x}{2}=0,-2 \sin 5 x \sin 2 x+2 \sin 5 x \cos 3 x=0$,

$-2 \sin 5 x(\sin 2 x-\cos 3 x)=0$.

From this:

1) $\sin 5 x=0,5 x=\pi n, x_{1}=\frac{\pi n}{5}, n \in Z$
2) $\sin 2 x-\cos 3 x=0 \Leftrightarrow \sin 2 x-\sin \left(\frac{\pi}{2}-3 x\right) \Leftrightarrow$

$\Leftrightarrow 2 \cos \frac{2 x+\frac{\pi}{2}-3 x}{2} \sin \frac{2 x-\frac{\pi}{2}+3 x}{2}=0,2 \cos \left(\frac{\pi}{4}-\frac{x}{2}\right) \sin \left(\frac{5}{2} x-\frac{\pi}{4}\right)=0$.

From this
a) $\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)=\cos \left(\frac{x}{2}-\frac{\pi}{4}\right)=0, \frac{x}{2}-\frac{\pi}{4}=\frac{\pi}{2}+\pi k, x_{2}=\frac{3}{2} \pi+2 \pi k=$

$=-\frac{\pi}{2}+2 \pi k=\frac{\pi}{2}(4 k-1), k \in Z$

b) $\sin \left(\frac{5}{2} x-\frac{\pi}{4}\right)=0, \frac{5}{2} x-\frac{\pi}{4}=\pi l, x_{3}=\frac{\pi}{10}+\frac{2}{5} \pi l=\frac{\pi}{10}(4 l+1), l \in Z$.

Answer: $x_{1}=\frac{\pi n}{5} ; x_{2}=\frac{\pi}{2}(4 k-1) ; x_{3}=\frac{\pi}{10}(4 l+1), n, k, l \in Z$.