Question

помогите решить пожалуйста​

Answer 1

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

Ответ:  №2.