Question

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$\sqrt{3} sinx-cosx=-\sqrt{2}$

Answer 1

$\sqrt3sinx-cosx=-\sqrt2\; |:2\\\\\frac{\sqrt3}{2}\cdot sinx-\frac{1}{2}\cdot cosx=-\frac{\sqrt2}{2}\\\\\frac{\sqrt3}{2}=cos\frac{\pi}{6}\; ,\; \; \frac{1}{2}=sin\frac{\pi}{6}\\\\sinx\cdot cos\frac{\pi }{6}-cosx\cdot sin\frac{\pi }{6}=-\frac{\sqrt2}{2}\\\\sin(x-\frac{\pi }{6})=-\frac{\sqrt2}{2}\\\\x-\frac{\pi }{6}=(-1)^{n}\cdot arcsin(-\frac{\sqrt2}{2})+\pi n=(-1)^{n+1}\cdot \frac{\pi }{4}+\pi n,\; n\in Z\\\\x=\frac{\pi }{6}+(-1)^{n+1}\cdot \frac{\pi }{4}+\pi n,\; n\in Z$