8.023. $\sin 7 x+\sin 9 x=2\left(\cos ^{2}\left(\frac{\pi}{4}-x\right)-\cos ^{2}\left(\frac{\pi}{4}+2 x\right)\right)$

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8.023. $\sin 7 x+\sin 9 x=2\left(\cos ^{2}\left(\frac{\pi}{4}-x\right)-\cos ^{2}\left(\frac{\pi}{4}+2 x\right)\right)$
## Solution.

From the condition

$2 \sin 8 x \cos x=1+\cos \left(\frac{\pi}{2}-2 x\right)-1-\cos \left(\frac{\pi}{2}+4 x\right) \Leftrightarrow$

$\Leftrightarrow 2 \sin 8 x \cos x=\sin 2 x+\sin 4 x \Leftrightarrow$

$\Leftrightarrow 2 \sin 8 x \cos x-2 \sin 3 x \cos x=0, \Leftrightarrow \cos x(\sin 8 x-\sin 3 x)=0$.

Then:

1) $\cos x=0$ or
2) $\sin 8 x-\sin 3 x=0$,
from which 1) $\left.x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1) \quad k \in Z, 2\right) 2 \sin \frac{5}{2} x \cos \frac{11}{2} x=0, \sin \frac{5}{2} . x=0$, $\frac{5}{2} x=\pi n, x_{2}=\frac{2}{5} \pi n, n \in Z ; \cos \frac{11}{2} x=0, \frac{11}{2} x=\frac{\pi}{2}+\pi l, x_{3}=\frac{\pi}{11}+\frac{2}{11} \pi l=$ $=\frac{\pi}{11}(2 l+1), \quad l \in Z$.

Answer: $\quad x_{1}=\frac{\pi}{2}(2 k+1) x_{2}=\frac{2}{5} \pi n ; x_{3}=\frac{\pi}{11}(2 l+1), k, n, l \in Z$.