The number of solutions of equation $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$

$8\sin x\equiv \frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$

$\Rightarrow 8{\sin }^{2}x\cos x=\sqrt{3}\sin x+\cos x$

$4(1-\cos 2x)\cos x=\sqrt{3}\sin x+\cos x$

$3\cos x-2(\cos 3x+\cos x)=\sqrt{3}\sin x$

$\cos x-2\cos 3x=\sqrt{3}\sin x$

$\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x=\cos 3x$

$\cos \left(x+\frac{\pi }{3}\right)=\cos 3x$

$$\begin{gathered} 3x=2nx\pm \left(x+\frac{\pi }{3}\right) \\ (+) \end{gathered}$$

$3x=2x\pi +x+\frac{\pi }{3}$

$2x=2n\pi +\frac{\pi }{3}$

$x=nx+\frac{\pi }{6}$

$[0,2\pi ]$

$\frac{\pi }{6}$

$$\begin{gathered} \frac{5\pi }{6} \\ (-) \end{gathered}$$

$3x=2n\pi -x-\frac{\pi }{3}$

$4x=2n\pi -\frac{\pi }{3}$

$x=\frac{n\pi }{2}-\frac{\pi }{12}$

$\frac{\pi }{2}-\frac{\pi }{12}=\frac{6\pi -\pi }{12}=\frac{5\pi }{12}$

$\pi -\frac{\pi }{12}=\frac{11\pi }{12}$

$\frac{3\pi }{2}-\frac{\pi }{12}=\frac{18\pi -\pi }{12}=\frac{17\pi }{12}$

$2\pi -\frac{\pi }{12}=\frac{23\pi }{12}$