Solve for x : $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$ 
Solve $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$ for $x$

I started by multiplying both sides of the equation by $\frac{1}{2\sqrt{2}}$ to obtain
$$\displaystyle\frac{\sin(x)}{2}+\frac{\sqrt{3}\cos(x)}{2} = \frac{\sqrt{3} +1}{2\sqrt{2}}$$ $$\iff \sin(60+x) = \frac{\sqrt{3} +1}{2\sqrt{2}}$$
I am stuck here. Any hints on solving the R.H.S will be appreciated.

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}

