8.374. $12 \sin x + 4 \sqrt{3} \cos (\pi + x) = m \sqrt{3}$.
## Solution.

From the condition we have

$$
\begin{aligned}
& 12 \sin x-4 \sqrt{3} \cos x=8 \sqrt{3}\left(\frac{\sqrt{3}}{2} \sin x-\frac{1}{2} \cos x\right)=8 \sqrt{3}\left(\sin 60^{\circ} \sin x-\right. \\
& \left.-\cos 60^{\circ} \cos x\right)=-8 \sqrt{3} \cos \left(60^{\circ}+x\right)=8 \sqrt{3} \sin \left(x-30^{\circ}\right) \\
& \text { then }
\end{aligned}
$$

$$
\begin{aligned}
& 8 \sqrt{3} \sin \left(x-30^{\circ}\right)=m \sqrt{3}, \sin \left(x-\frac{\pi}{6}\right)=\frac{m}{8} \\
& x=(-1)^{n} \arcsin \frac{m}{8}+\pi n+\frac{\pi}{6}=(-1)^{n} \arcsin \frac{m}{8}+\frac{\pi}{6}(6 n+1)
\end{aligned}
$$

where $-8 \leq m \leq 8, n \in Z$.

Answer: $x=(-1)^{n} \arcsin \frac{m}{8}+\frac{\pi}{6}(6 n+1)$, where $-8 \leq m \leq 8, n \in Z$.