Giải phương trình: \(\cos x - \sqrt 3 \sin x = 2\cos 3x.\)
Giải chi tiết:
Chia cả 2 vế cho \(\sqrt {1 + {{\left( {\sqrt 3 } \right)}^2}} = 2\), ta có:
\(\begin{array}{l}\frac{1}{2}.\cos x - \frac{{\sqrt 3 }}{2}.\sin x = \cos 3x \Leftrightarrow \cos \left( {x + \frac{\pi }{3}} \right) = \cos 3x\\ \Leftrightarrow \left[ \begin{array}{l}x + \frac{\pi }{3} = 3x + k2\pi \\x + \frac{\pi }{3} = - 3x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} - k\pi \\x = \frac{\pi }{{12}} + \frac{{k\pi }}{2}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
KL: \(x \in \left\{ {\frac{\pi }{6} - k\pi ;\frac{\pi }{{12}} + \frac{{k\pi }}{2};k \in \mathbb{Z}} \right\}\) .
Chọn C.