Ans: C
$\begin{array}{rcl}
\sin^2x & = & 6\cos^2 x \\
\left(\dfrac{\sin x}{\cos x}\right)^2 & = & 6 \\
\tan x & = & \pm \sqrt{6} \\
\end{array}$
$\therefore \tan x = \sqrt{6}$ or $\tan x = -\sqrt{6}$.
For $\tan x = \sqrt{6}$, $x=67.792\ 345\ 7^\circ$ or $247.792\ 345\ 7^\circ$.
For $\tan x = -\sqrt{6}$, $x=112.207\ 654\ 3^\circ$ or $292.207\ 654\ 3^\circ$.
Therefore, there are $4$ roots.
