Giải phương trình lượng giác: a) 2 cos x − 1 = 0 b) sin 3 x = cos 2 x

a) \[2\cos x - 1 = 0 \Leftrightarrow \cos x = \frac{1}{2}\]

\[x = \pm \frac{\pi }{3} + k2\pi \]

b) \[\sin 3x = \cos 2x \Leftrightarrow \sin 3x = \sin \left( {\frac{\pi }{2} - 2x} \right)\]

\[ \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{2} - 2x + k2\pi \\3x = \pi - \frac{\pi }{2} + 2x + k2\pi \end{array} \right.\]\[ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} + k2\pi \\x = \frac{\pi }{2} + k2\pi \end{array} \right.\]\[ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + k\frac{{2\pi }}{5}\\x = \frac{\pi }{2} + k2\pi \end{array} \right.\]