Đáp án:
$\left[\begin{array}{l} x=\dfrac{\pi}{6}+k \pi(k \in \mathbb{Z})\\ x=\dfrac{\pi}{2}+k\pi(k \in \mathbb{Z})\end{array} \right..$
Giải thích các bước giải:
$2\sin^2x+\sqrt{3}\sin 2x=2\\ \Leftrightarrow 2.\dfrac{1-\cos 2x}{2}+\sqrt{3}\sin 2x=2\\ \Leftrightarrow 1-\cos 2x+\sqrt{3}\sin 2x=2\\ \Leftrightarrow \sqrt{3}\sin 2x-\cos 2x=1\\ \Leftrightarrow \dfrac{\sqrt{3}}{2}\sin 2x-\dfrac{1}{2}\cos 2x=\dfrac{1}{2}\\ \Leftrightarrow \sin 2x\cos\dfrac{\pi}{6}-\cos 2x\sin \dfrac{\pi}{6}=\dfrac{1}{2}\\ \Leftrightarrow \sin\left( 2x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\\ \Leftrightarrow \left[\begin{array}{l} 2x-\dfrac{\pi}{6}=\dfrac{\pi}{6}+k 2 \pi(k \in \mathbb{Z})\\ 2x-\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\\Leftrightarrow \left[\begin{array}{l} 2x=\dfrac{\pi}{3}+k 2 \pi(k \in \mathbb{Z})\\ 2x=\pi+k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\\Leftrightarrow \left[\begin{array}{l} x=\dfrac{\pi}{6}+k \pi(k \in \mathbb{Z})\\ x=\dfrac{\pi}{2}+k\pi(k \in \mathbb{Z})\end{array} \right..$
