square both sides, we get
$$2 \sin^2(x) + 6 \cos^2(x) + 2\sqrt{12}\cos(x)\sin(x) = (1 + \sqrt{3})^2$$
that is
$$2 + 4 \cos^2(x) + 2\sqrt{12}\cos(x)\sin(x) = (1 + \sqrt{3})^2$$
that is
$$2 + 4 \cos^2(x) + \sqrt{12}\sin(2x) = (1 + \sqrt{3})^2$$
that is
$$2 + 4 \frac{1}{1+\tan^2(x)} + \sqrt{12}\frac{2\tan x}{1+\tan^2(x)} = (1 + \sqrt{3})^2$$
Multiply everything by $1 + \tan^2(x)$
$$2(1+\tan^2(x)) + 4 +2 \sqrt{12}\tan x = (1 + \sqrt{3})^2(1+\tan^2(x)) $$
This is quadratic in $y = \tan(x)$
$$ay^2 + by + c = 0$$
where
\begin{align}
a &= 2- (1 + \sqrt{3})^2\\
b &=2\sqrt{12}\\
c &= 2 + 4 - (1 + \sqrt{3})^2
\end{align}
Solving you get something like
\begin{align}
y_1 &= 1.0000 \\
y_2 &= 0.2679
\end{align}
Now taking the inverse tangent of that you get
\begin{align}
x_1 &= \tan^{-1} y_1 =\tan^{-1} 1.0000 = 45^\circ\\
x_2 &= \tan^{-1} y_2 =\tan^{-1} 0.2679= 15^\circ
\end{align}