8.156. $(\sin 2 x+\sqrt{3} \cos 2 x)^{2}=2-2 \cos \left(\frac{2}{3} \pi-x\right)$.
Solution.

We have

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

From this:

1) $\sin \frac{5 x}{2}=0, \frac{5 x}{2}=\pi n, x_{1}=\frac{2}{5} \pi n, n \in Z$;
2) $\cos \left(\frac{3 x}{2}+\frac{\pi}{6}\right)=0, \frac{3 x}{2}=-\frac{\pi}{6}+\frac{\pi}{2}+\pi k, x_{2}=\frac{2}{9} \pi+\frac{2}{3} \pi k=\frac{2 \pi}{9}(3 k+1)$, $k \in Z$.

Answer: $x_{1}=\frac{2}{5} \pi n, x_{2}=\frac{2 \pi}{9}(3 k+1), n, k \in Z$.