8.048. $\sin \left(\frac{\pi}{4}+5 x\right) \cos \left(\frac{\pi}{4}+2 x\right)=\sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-6 x\right)$
## Solution.

Rewrite the given equation as

$$
\begin{aligned}
& \frac{1}{2}\left(\sin \left(\frac{\pi}{4}+5 x-\frac{\pi}{4}-2 x\right)+\sin \left(\frac{\pi}{4}+5 x+\frac{\pi}{4}+2 x\right)\right)= \\
& =\frac{1}{2}\left(\cos \left(\frac{\pi}{4}+x-\frac{\pi}{4}+6 x\right)-\cos \left(\frac{\pi}{4}+x+\frac{\pi}{4}-6 x\right)\right) \Leftrightarrow \sin 3 x+ \\
& +\sin \left(\frac{\pi}{2}+7 x\right)-\cos 7 x+\cos \left(\frac{\pi}{2}-5 x\right)=0 \Leftrightarrow \sin 3 x+\cos 7 x-\cos 7 x+ \\
& +\sin 5 x=0, \sin 3 x+\sin 5 x=0 \Leftrightarrow 2 \sin \frac{5 x+3 x}{2} \cos \frac{5 x-3 x}{2}=0, \\
& \sin 4 x \cos x=0
\end{aligned}
$$

From this:

1) $\sin 4 x=0, 4 x=\pi n, x_{1}=\frac{\pi n}{4}, n \in Z$
2) $\cos x=0, \quad x_{2}=\frac{\pi}{2}+\pi k, k \in Z ; x_{2}$ is included in $x_{1}$.

Answer: $x=\frac{\pi n}{4}, n \in Z$