Giải phương trình: \( \sqrt 3 \sin 2x + \cos 2x = 1 \)
Giải chi tiết:
\(\sqrt 3 \sin 2x + \cos 2x = 1 \Leftrightarrow \frac{{\sqrt 3 }}{2}\sin 2x + \frac{1}{2}\cos 2x = \frac{1}{2}\) (chia cả 2 vế cho 2).
\(\begin{array}{l} \Leftrightarrow \cos \left( {\frac{\pi }{6}} \right).\sin 2x + \sin \left( {\frac{\pi }{6}} \right).\cos 2x = \sin \left( {\frac{\pi }{6}} \right) \Leftrightarrow \sin \left( {2x + \frac{\pi }{6}} \right) = \sin \left( {\frac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x + \frac{\pi }{6} = \frac{\pi }{6} + k2\pi \\2x + \frac{\pi }{6} = \pi - \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{\pi }{3} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy \(S = \left\{ {k\pi ;\frac{\pi }{3} + k\pi } \right\}\).
Chọn C.