Đáp án:
$x=\dfrac{4\pi}{3}+k 2 \pi(k \in \mathbb{Z}).$
Giải thích các bước giải:
$\sqrt{3}\sin x+\cos x=-2\\ \Leftrightarrow \dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x=-1\\ \Leftrightarrow \sin\dfrac{\pi}{3}\sin x+\cos\dfrac{\pi}{3}\cos x=-1\\ \Leftrightarrow \cos x\cos\dfrac{\pi}{3}+\sin x\sin\dfrac{\pi}{3}=-1\\ \Leftrightarrow \cos \left(x-\dfrac{\pi}{3}\right)=-1\\ \Leftrightarrow x-\dfrac{\pi}{3}=\pi+k 2 \pi(k \in \mathbb{Z})\\ \Leftrightarrow x=\dfrac{4\pi}{3}+k 2 \pi(k \in \mathbb{Z}).$
