Giải phương trình lượng giác sau; 2sinx - √3 = 0

`~rai~`

\(2\sin x-\sqrt{3}=0\\\Leftrightarrow \sin x=\dfrac{\sqrt{3}}{2}\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{3}+k2\pi\\x=\pi-\dfrac{\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{\pi}{3}+k2\pi;\dfrac{2\pi}{3}+k2\pi\Big|k\in\mathbb{Z}\right\}.\)