If   ${\log }_{\tan x}\left(2+4{\cos }^{2}x\right)=2$  then $x=$

$2+4{\cos }^{2}x={\tan }^{2}x$

$2+4{\cos }^{2}x=\frac{{\sin }^{2}x}{{\cos }^{2}x}$

$4{\cos }^{4}x+3{\cos }^{2}x-1=0$

$\left(4{\cos }^{2}x-1\right)\left({\cos }^{2}x+1\right)=0$

${\cos }^{2}x=\frac{1}{4}\Rightarrow x=n\pi \pm \frac{\pi }{3}$

But for $x=n\pi -\frac{\pi }{3}$the equation ${\log }_{\tan x}\left(2+4{\cos }^{2}x\right)=2$ is not defined since the base tan x becomes negataive