Tìm nghiệm phương trình lượng giác sin ( x − 2 π /3 ) − cos 2x = 0 ;

\(\begin{array}{l}\sin \left( {x - \frac{{2\pi }}{3}} \right) - \cos 2x = 0 \Leftrightarrow \sin \left( {x - \frac{{2\pi }}{3}} \right) = \cos 2x\\ \Leftrightarrow \sin \left( {x - \frac{{2\pi }}{3}} \right) = \sin \left( {\frac{\pi }{2} - 2x} \right) \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x - \frac{{2\pi }}{3} = \frac{\pi }{2} - 2x + k2\pi }\\{x - \frac{{2\pi }}{3} = \pi  - \frac{\pi }{2} + 2x + k2\pi }\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{3x - \frac{{2\pi }}{3} = \frac{\pi }{2} + k2\pi }\\{ - x - \frac{{2\pi }}{3} = \frac{\pi }{2} + k2\pi }\end{array} \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = \frac{{7\pi }}{{18}} + \frac{{k2\pi }}{3}}\\{x =  - \frac{{7\pi }}{6} - k2\pi }\end{array},k \in \mathbb{Z}} \right.} \right.\end{array}\)