2022-II-39

$\begin{array}{rcl}{\sin }^{2}x& =& 6{\cos }^{2}x\\ {\left({\displaystyle \frac{\sin x}{\cos x}}\right)}^2& =& 6\\ \tan x& =& \pm \sqrt{6}\end{array}$

$\therefore \tan x=\sqrt{6}$ or $\tan x=-\sqrt{6}$.

For $\tan x=\sqrt{6}$, $x=67.7923457^{\circ }$ or $247.7923457^{\circ }$.

For $\tan x=-\sqrt{6}$, $x=112.2076543^{\circ }$ or $292.2076543^{\circ }$.

Therefore, there are $4$ roots.