$\sin 2x-\frac{\sin \frac{\pi }{3}}{\cos \frac{\pi }{3}}\cos 2x=\sqrt{3}\Rightarrow \frac{\sin 2x\cos \frac{\pi }{3}-\cos 2x\sin \frac{\pi }{3}}{\cos \frac{\pi }{3}}=\sqrt{3}$  $\sin \left( 2x-\frac{\pi }{3} \right)=\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3}$  $\left\{ \begin{matrix}    2x-\frac{\pi }{3}=2k\pi +\frac{\pi }{3}\Rightarrow x=k\pi +\frac{\pi }{3}  \\    2x-\frac{\pi }{3}=2k\pi +\pi -\frac{\pi }{3}\Rightarrow x=k\pi +\frac{\pi }{2}  \\ \end{matrix},\left( k\in z \right)\xrightarrow{x\in \left[ \frac{\pi }{2},\frac{3\pi }{2} \right]}x=\frac{4\pi }{3},\frac{\pi }{2},\frac{3\pi }{2} \right.$