An elegant way to solve $\frac {\sqrt3 - 1}{\sin x} + \frac {\sqrt3 + 1}{\cos x} = 4\sqrt2 $ The question  is to find $x\in\left(0,\frac{\pi}{2}\right)$:
$$\frac {\sqrt3 - 1}{\sin x}  + \frac {\sqrt3  + 1}{\cos x} = 4\sqrt2 $$
What I did was to take the $\cos x$  fraction    to the right and try to simplify ;
But it looked very messy and trying to write $\sin x$  in terms of $\cos x$ didn't help. 
Is there a more simple (elegant) way to do this. 
 A: Another approach is to write the equation as
$(\sqrt{3} - 1) \cos x +(\sqrt{3} + 1) \sin x = 4 \sqrt2 \sin x \cos x$
then rearranging gives
$\frac{\sqrt{3} - 1}{2 \sqrt{2}} \cos x +\frac{\sqrt{3} + 1}{2 \sqrt{2}} \sin x = 2  \sin x \cos x$.
Now note that $(\frac{\sqrt{3} - 1}{2 \sqrt{2}})^2 +(\frac{\sqrt{3} + 1}{2 \sqrt{2}})^2 = \frac{1}{8} ((3 -2 \sqrt{3} + 1) + (3 +2 \sqrt{3} + 1) = 1$ so that we can write $\frac{\sqrt{3} - 1}{2 \sqrt{2}}$ as the $\sin$ of some angle, say $\alpha$, and identify $\frac{\sqrt{3} + 1}{2 \sqrt{2}}$ as $\cos \alpha$.
Then you have to solve $\sin(\alpha + x) = \sin(2x)$, with $\tan \alpha = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$.
A: Replace $\sqrt3=\tan60^\circ,1=\tan45^\circ$
to find $$\sin(15^\circ+x)=2\sin x\cos x=\sin2x$$
Now if $\sin y=\sin A, y=n180^\circ+(-1)^nA$ where $n$ is any integer
As $\sqrt3-1=\tan60^\circ-\tan45^\circ=\dfrac{\sin(60-45)^\circ}{\cos60^\circ\cos45^\circ}=2\sqrt2\sin15^\circ$
Similarly, $\sqrt3+1=2\sqrt2\sin75^\circ=2\sqrt2\cos15^\circ$
A: $\sin(\frac{\pi}{12}) = \frac{\sqrt{3}-1}{2\sqrt{2}}$ and $\cos(\frac{\pi}{12}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$. Plugging them into your equation yields $\sin(x+\frac{\pi}{12}) = \sin(2x)$. So $x = \frac{\pi}{12}$
