\(\begin{array}{l}\cos 2 x+7 \cos x-\sqrt{3}(\sin 2 x-7 \sin x)=8 \Leftrightarrow \cos 2 x-\sqrt{3} \sin 2 x+7(\cos x+\sqrt{3} \sin x)=8 \\ \Leftrightarrow \frac{1}{2} \cos 2 x-\frac{\sqrt{3}}{2} \sin 2 x+7\left(\frac{1}{2} \cos x+\frac{\sqrt{3}}{2} \sin x\right)=4 \Leftrightarrow \cos \left(2 x+\frac{\pi}{3}\right)+7 \cos \left(x-\frac{\pi}{3}\right)=4 \\ \Leftrightarrow \cos \left(2\left(x+\frac{\pi}{6}\right)\right)-7 \sin \left(x+\frac{\pi}{6}\right)=4 \Leftrightarrow 1-2 \sin ^{2}\left(x+\frac{\pi}{6}\right)-7 \sin \left(x+\frac{\pi}{6}\right)=4 \\ \Leftrightarrow 2 \sin ^{2}\left(x+\frac{\pi}{6}\right)+7 \sin \left(x+\frac{\pi}{6}\right)+3=0 \Leftrightarrow\left[\begin{array}{l}\sin \left(x+\frac{\pi}{6}\right)=-\frac{1}{2} \\ \sin \left(x+\frac{\pi}{6}\right)=-3 \text { (Vô nghiẹm) }\end{array}\right. \\ \text { +) } \sin \left(x+\frac{\pi}{6}\right)=-\frac{1}{2} \Leftrightarrow \sin \left(x+\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right) \Leftrightarrow\left[\begin{array}{l}x=-\frac{\pi}{3}+k 2 \pi \\ x=\pi+k 2 \pi\end{array},(k \in \mathbb{Z}) .\right.\end{array}\)