a) Cho biết \(\sin x = \frac{3}{4}.\) Tính giá trị của biểu thức \(P = {\sin ^2}2x.\)

b) Giải phương trình \(\sin 2x - \cos \left( {x - \frac{\pi }{6}} \right) = 0.\)

a) Ta có: \({\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x = 1 - {\left( {\frac{3}{4}} \right)^2} = \frac{7}{{16}}.\)

\( \Rightarrow P = {\sin ^2}2x = {\left( {2\sin x.\cos x} \right)^2} = 4{\sin ^2}x.{\cos ^2}x = 4.{\left( {\frac{3}{4}} \right)^2}.\frac{7}{{16}} = \frac{{63}}{{64}}.\)

b) Ta có: \(\sin 2x - \cos \left( {x - \frac{\pi }{6}} \right) = 0 \Leftrightarrow \sin 2x = \cos \left( {x - \frac{\pi }{6}} \right)\)

\( \Leftrightarrow \cos \left( {\frac{\pi }{2} - 2x} \right) = \cos \left( {x - \frac{\pi }{6}} \right) \Leftrightarrow \left[ \begin{array}{l}\frac{\pi }{2} - 2x = x - \frac{\pi }{6} + k2\pi \\\frac{\pi }{2} - 2x = - \left( {x - \frac{\pi }{6}} \right) + k2\pi \end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{{2\pi }}{9} - \frac{{k2\pi }}{3}\\x = \frac{\pi }{3} - k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right).\)