<s>[INST] solve a trigonometric equation $\sqrt{3} \sin(x)-\cos(x)=\sqrt{2}$ $$\sqrt{3}\sin{x} - \cos{x} = \sqrt{2} $$
I think to do :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{2}}$$
but i dont get anything.
Or to divied by $\sqrt{3}$ :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{3}}$$
 [/INST] Hint:
$$\sqrt3\sin x-\cos x=\sqrt2\iff \sin\frac\pi3\sin x-\cos\frac\pi3\cos x=\frac{\sqrt2}2$$
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