General solution of the equation $\sec 2x=\frac{-2}{\sqrt{3}}$ is

The correct option is B $n\pi \pm \frac{5\pi }{12},n\in Z$
$\sec 2x=\frac{-2}{\sqrt{3}}\Rightarrow \cos 2x=\frac{-\sqrt{3}}{2}\Rightarrow \cos 2x=\cos \frac{5\pi }{6}\Rightarrow 2x=2n\pi \pm \frac{5\pi }{6}\Rightarrow x=n\pi \pm \frac{5\pi }{12},n\in Z$