$8.423 \sqrt[3]{\sin ^{2} x}-\sqrt[3]{\cos ^{2} x}=\sqrt[3]{2 \cos 2 x}$.
Solution. The original equation is equivalent to the following:

$$
\left(\sqrt[3]{\sin ^{2} x}-\sqrt[3]{\cos ^{2} x}\right)^{3}=2\left(\cos ^{2} x-\sin ^{2} x\right)
$$

Let: $u=\sqrt[3]{\sin ^{2} x}, v=\sqrt[3]{\cos ^{2} x}$. Then

$$
(u-v)^{3}=2\left(v^{3}-u^{3}\right) \Leftrightarrow(u-v)\left(u^{2}+v^{2}\right)=0 \Leftrightarrow u=v \Leftrightarrow\left[\begin{array}{l}
\sin x=\cos x, \\
\sin x=-\cos x
\end{array} \Leftrightarrow x=\frac{\pi}{4}+\frac{\pi k}{2}\right.
$$

Answer: $\frac{\pi}{4}(2 k+1)$.