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.


$\sin 75^\text o= \dfrac{\sqrt{3}+1}{2\sqrt2}$

Can you find now ?
PROOF :

$\sin (x+y)  = \sin x\cos y +\cos x \sin y  $
$\sin(75^\text o) = \sin(30^\text o+45^\text o)\\\qquad\quad= \sin 30^\text o\cos 45^\text o +\cos 30^\text o \sin 45^\text o\\\qquad\quad = \dfrac{1}{2\sqrt2}+\dfrac{\sqrt3}{2\sqrt2} = \dfrac{\sqrt3+1}{2\sqrt2} $

