ĐK: $\sin x\ne 0; \cos x\ne 0$
$\to \sin2x=2\sin x\cos x\ne 0$
$\to x\ne \dfrac{k\pi}{2}$
$8\sin x=\dfrac{\sqrt3}{\cos x}+\dfrac{1}{\sin x}$
$\to 8\sin x\cos x\sin x=\sqrt3\sin x+\cos x$
$\to 4\sin2x.\sin x=\sqrt3\sin x+\cos x$
$\to -2(\cos3x-\cos x)=\sqrt3\sin x+\cos x$
$\to -2\cos3x+2\cos x=\sqrt3\sin x+\cos x$
$\to -2\cos3x=\sqrt3\sin x-\cos x$
$\to -2\cos3x=2\sin\left(x-\dfrac{\pi}{6}\right)$
$\to \sin\left(3x-\dfrac{\pi}{2}\right)=\sin\left(x-\dfrac{\pi}{6}\right)$
$\to \left[ \begin{array}{l}3x-\dfrac{\pi}{2}= x-\dfrac{\pi}{6}+k2\pi \\3x-\dfrac{\pi}{2}= \pi+\dfrac{\pi}{6}-x+k2\pi\end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{5\pi}{12}+\dfrac{k\pi}{2}\end{array} \right.$ (TM)
