<s>[INST] 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.
 [/INST] Try to find what is $$\sin\frac{5\pi}{12}$$
If you get it $$\frac{\sqrt3+1}{2\sqrt2}$$ then you did correctly and you know what to do next using the general definition of 
$$\sin x=\sin y $$ yields what you know it !
you can find $$\sin\frac{5\pi}{12}$$ using the formula for $$\sin \frac{x}{2}$$ by taking $x=\frac{5\pi}{6}$ in degrees which is equivalent to 75 degrees
</s>