8.051. $1+\sin 2 x=(\cos 3 x+\sin 3 x)^{2}$.
## Solution.

We have

$$
\begin{aligned}
& 1+\sin 2 x=\cos ^{2} 3 x+2 \sin 3 x \cos 3 x+\sin ^{2} 3 x, \quad 1+\sin 2 x=1+\sin 6 x \\
& \sin 6 x-\sin 2 x=0 \Leftrightarrow 2 \cos \frac{6 x+2 x}{2} \sin \frac{6 x-2 x}{2}=0, \quad \cos 4 x \sin 2 x=0
\end{aligned}
$$

from which:

1) $\cos 4 x=0, \quad 4 x=\frac{\pi}{2}+\pi k, \quad x_{1}=\frac{\pi}{8}+\frac{\pi k}{4}=\frac{\pi}{8}(2 k+1), \quad k \in Z$;
2) $\sin 2 x=0,2 x=\pi n, x_{2}=\frac{\pi n}{2}, n \in Z$.

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