Solving $2 \sin x \left( \sqrt{3} \cos x- \sin x \right)= \sqrt{2} - 1$ I wanted to know if this equation could be solved any further please.
$$2 \sin x \left( \sqrt{3} \cos x- \sin x \right)= \sqrt{2} - 1$$
I have gone this far:
$$4 \sin x \sin(60^\circ-x)= \sqrt{2}- 1$$
Thank you

Hint: $2\sin x(\sqrt{3}\cos x - \sin x) = \sqrt{2} - 1\implies \sqrt{3}\sin(2x) - (1- \cos(2x)) = \sqrt{2} - 1\implies \sqrt{3}\sin(2x)+\cos(2x)=\sqrt{2}\implies \sin(2x+\frac{\pi}{6})= \sin(\frac{\pi}{4})$. At this point, can you continue to the finish line ….?
