Giải phương trình: \(\sin 3x - \sqrt 3 \cos 3x = 2\sin 2x.\)
Giải chi tiết:
Chia cả 2 vế cho \(\sqrt {1 + {{\left( {\sqrt 3 } \right)}^2}} = 2\) , ta có:
\(\begin{array}{l}\frac{1}{2}.\sin 3x - \frac{{\sqrt 3 }}{2}.\cos 3x = \sin 2x \Leftrightarrow \sin \left( {3x - \frac{\pi }{3}} \right) = \sin 2x\\ \Leftrightarrow \left[ \begin{array}{l}3x - \frac{\pi }{3} = 2x + k2\pi \\3x - \frac{\pi }{3} = \pi - 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = \frac{{4\pi }}{{15}} + \frac{{k2\pi }}{5}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
KL: \(x \in \left\{ {\frac{\pi }{3} + k2\pi ;\frac{{4\pi }}{{15}} + \frac{{k2\pi }}{5},k \in \mathbb{Z}} \right\}\).
Chọn B.