8.020. $1-\sin 3 x=\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^{2}$.

8.020. $1-\sin 3x=\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^{2}$.
## Solution.

## Given

$1-\sin 3 x=\sin ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}+\cos ^{2} \cdot \frac{x}{2}, \quad \sin 3 x-\sin x=0 \Leftrightarrow$

$\Leftrightarrow 2 \sin x \cos 2 x=0$.

Then either

1) $\sin x=0, \quad x_{1}=\pi n, \quad n \in Z$,

## or

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

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