8.013. $\sin 2 x=\cos ^{4} \frac{x}{2}-\sin ^{4} \frac{x}{2$.

8.013. $\sin 2 x=\cos ^{4} \frac{x}{2}-\sin ^{4} \frac{x}{2}$. 

(Note: The original text had a minor formatting issue in the equation, which has been corrected in the translation.)
Solution.

We have

$2 \sin x \cos x-\left(\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}\right)\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)=0$,

$2 \sin x \cos x-\cos x=0 \Leftrightarrow \cos x(2 \sin x-1)=0$.

Then:

1) $\cos x=0, x_{1}=\frac{\pi}{2}+\pi n=\frac{\pi}{2}(2 n+1), n \in Z$;

or
2) $2 \sin x-1=0, \quad \sin x=\frac{1}{2}, \quad x_{2}=(-1)^{k} \frac{\pi}{6}+\pi k, \quad k \in Z$.

Answer: $\quad x_{1}=\frac{\pi}{2}(2 n+1) x_{2}=(-1)^{k} \frac{\pi}{6}+\pi k, \quad n, k \in Z$.