<s>[INST] How to find the roots of $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$? Find all $x$ in the interval $(0,\pi/2)$ such that $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$.
The options are (i)$\pi/9,2\pi/7$, (ii)$\pi/36,11\pi/12$ (iii)$\pi/12,11\pi/36$ (iv) All
I have been able to find one value of $x$, $\pi/12$.  How do I find the other root(s)? 
My attempt:
$\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$
or, $\frac{\sin\pi/3-\sin\pi/6}{\sin x}+\frac{\cos\pi/6+\cos\pi/3}{\cos x}=2\sqrt{2}$
or, $\frac{\sin(\pi/4)cos(\pi/12)}{\sin x}+\frac{\cos(\pi/4)cos(\pi/12)}{\cos x}=\sqrt{2}$
or, $\sin(x+\pi/12)=\sin2x$
or, $x=\pi/12$
 [/INST] Use $$\sin15^{\circ}=\frac{\sqrt3-1}{2\sqrt2}$$ and
$$\cos15^{\circ}=\frac{\sqrt3+1}{2\sqrt2}.$$
We obtain:
$$\sin(15^{\circ}+x)=\sin2x.$$
Thus, $$15^{\circ}+x=2x+360^{\circ}k,$$ where $k$ is an integer number, or
$$15^{\circ}+x=180^{\circ}-2x+360^{\circ}k.$$
Can you end it now?
</s>