Đáp án:
$C.$
Giải thích các bước giải:
$\cos^2x-\sqrt{3}\sin 2x=1+\sin^2x\\\Leftrightarrow 1-\sin^2x-\sqrt{3}\sin 2x=1+\sin^2x\\\Leftrightarrow 2\sin^2x+\sqrt{3}\sin 2x=0\\\Leftrightarrow 2\sin^2x+2\sqrt{3}\sin x\cos x=0\\\Leftrightarrow \sin^2x+\sqrt{3}\sin x\cos x=0\\\Leftrightarrow \sin x(\sin x+\sqrt{3}\cos x)=0\\\Leftrightarrow \sin x\left(\dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x\right)=0\\\Leftrightarrow \sin x\left(\sin x\cos \dfrac{\pi}{3}+\cos x\sin \dfrac{\pi}{3}\right)=0\\\Leftrightarrow \sin x.\sin \left(x+\dfrac{\pi}{3}\right)=0\\\Leftrightarrow \left[\begin{array}{l} \sin x=0\\\sin \left(x+\dfrac{\pi}{3}\right)=0\end{array} \right.\\\Leftrightarrow \left[\begin{array}{l} x=k \pi(k \in \mathbb{Z})\\x+\dfrac{\pi}{3}=l \pi(l \in \mathbb{Z})\end{array} \right.\\\Leftrightarrow \left[\begin{array}{l} x=k \pi(k \in \mathbb{Z})\\x=-\dfrac{\pi}{3}+l \pi(l \in \mathbb{Z})\end{array} \right.\\x \in (\pi;\pi) \Rightarrow \left[\begin{array}{l} -\pi<k \pi<\pi(k \in \mathbb{Z})\\-\pi<-\dfrac{\pi}{3}+l \pi<\pi(l \in \mathbb{Z})\end{array} \right.\\\Leftrightarrow \left[\begin{array}{l} -1<k <1(k \in \mathbb{Z})\\-\dfrac{2\pi}{3}<l \pi<\dfrac{4\pi}{3}(l \in \mathbb{Z})\end{array} \right.\\\Leftrightarrow \left[\begin{array}{l} -1<k <1(k \in \mathbb{Z})\\-\dfrac{2}{3}<l <\dfrac{4}{3}(l \in \mathbb{Z})\end{array} \right.\\\Leftrightarrow \left[\begin{array}{l} k=0\\l=0;l=1\end{array} \right.\\\Rightarrow x \in \left\{-\dfrac{\pi}{3};0;\dfrac{2\pi}{3}\right\}.$
