Phương trình \(2{\sin ^2}x + \sqrt 3 \sin 2x = 3\) có nghiệm là:
Giải chi tiết:
\(\begin{array}{l}\,\,\,\,2{\sin ^2}x + \sqrt 3 \sin 2x = 3 \Leftrightarrow 1 - \cos 2x + \sqrt 3 \sin 2x = 3\\ \Leftrightarrow \sqrt 3 \sin 2x - \cos 2x = 2 \Leftrightarrow \frac{{\sqrt 3 }}{2}\sin 2x - \frac{1}{2}\cos 2x = 1\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \frac{\pi }{2} \Leftrightarrow 2x - \frac{\pi }{6} = \frac{\pi }{2} + k2\pi \\ \Leftrightarrow 2x = \frac{{2\pi }}{3} + k2\pi \Leftrightarrow x = \frac{\pi }{3} + k\pi \,\,\,\,\,\,\,\,\left( {k \in \mathbb{Z}} \right).\end{array}\)
Chọn A.