If $6{\tan }^{2}x-2{\cos }^{2}x=\cos 2x$ then

The correct option is D ($\cos 2x=\frac{1}{2}$)

Given $6{\tan }^{2}x-2{\cos }^{2}x=\cos 2x$

$\Rightarrow 6\cdot \frac{{\sin }^{2}x}{{\cos }^{2}x}=\cos 2x+2{\cos }^{2}x$
Since $2{\sin }^{2}x=1-\cos 2x$ and $2{\cos }^{2}x=1+\cos 2x$

$\Rightarrow 6\cdot \frac{1-\cos 2x}{1+\cos 2x}=\cos 2x+(1+\cos 2x)$,

Put $\cos 2x=y$

$\Rightarrow 6\cdot \frac{1-y}{1+y}=2y+1$

$\Rightarrow 6(1-y)=(2y+1)(y+1)$

$\Rightarrow 6-6y=2y^2+2y+y+1$

$\Rightarrow 2y^2+9y-5=0$
$\Rightarrow y=\frac{-9\pm \sqrt{81+40}}{4}$

$\Rightarrow y=\frac{-9\pm \sqrt{121}}{4}$
$\Rightarrow y=\frac{-9\pm 11}{4}$
$=\frac{2}{4},\frac{-20}{4}$
$\therefore y=\frac{1}{2},-5$

$\Rightarrow \cos 2x=\frac{1}{2}$ or $\cos 2x=-5$(rejected)
$\therefore \cos 2x=\frac{1}{2}$