Giải phương trình sin (pi/2 + 2x) + căn bậc hai 3 sin (pi - 2x) = 1

\(\sin \left( {\frac{\pi }{2} + 2x} \right) + \sqrt 3 \sin \left( {\pi - 2x} \right) = 1\)

\(\begin{array}{l} \Leftrightarrow \cos 2x + \sqrt 3 \sin 2x = 1\\ \Leftrightarrow \sin \frac{\pi }{6}\cos 2x + \cos \frac{\pi }{6}.\sin 2x = \frac{1}{2}\end{array}\)

\(\begin{array}{l} \Leftrightarrow \sin \left( {2x + \frac{\pi }{6}} \right) = \frac{1}{2}\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x + \frac{\pi }{6} = \frac{\pi }{6} + k2\pi }\\{2x + \frac{\pi }{6} = \frac{{5\pi }}{6} + k2\pi }\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = k\pi }\\{x = \frac{\pi }{3} + k\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)