Đáp án:
$S=\left\{\dfrac{7\pi}{6}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\sqrt 3\cos x+\sin x=-2$
$⇔\dfrac{\sqrt 3}{2}\cos x+\dfrac{1}{2}\sin x=-1$
$⇔\cos x.\cos \dfrac{\pi}{6}+\sin x.\sin\dfrac{\pi}{6}=-1$
$⇔\cos\left(x-\dfrac{\pi}{6}\right)=-1$
$⇔x-\dfrac{\pi}{6}=\pi+k2\pi\,\,(k\in\mathbb Z)$
$⇔x=\dfrac{7\pi}{6}+k2\pi\,\,(k\in\mathbb Z)$
Vậy $S=\left\{\dfrac{7\pi}{6}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$.
