Đề bài: (Câu hỏi của bạn Huy Ninh hỏi)
Giải PTLG: $\ cos\,4x = co{s^2}3x + {\sin ^2}x.$
Giải:
\[cos\,4x = co{s^2}3x + {\sin ^2}x \Leftrightarrow cos\,4x = \frac{{1 + cos\,6x}}{2} + \frac{{1 - cos\,2x}}{2}\]\[ \Leftrightarrow cos\,4x = 1 + \frac{1}{2}\left( {cos\,6x - cos\,2x} \right) \Leftrightarrow cos\,4x = 1 - \sin 4x\sin 2x\]\[ \Leftrightarrow 2co{s^2}{\mkern 1mu} 2x - 1 = 1 - 2{\sin ^2}2x.cos{\mkern 1mu} 2x \Leftrightarrow 2co{s^2}{\mkern 1mu} 2x - 1 = 1 + 2cos{\mkern 1mu} 2x\left( {1 - co{s^2}{\mkern 1mu} 2x} \right)\]\[ \Leftrightarrow co{s^3}{\mkern 1mu} 2x + co{s^2}{\mkern 1mu} 2x - cos{\mkern 1mu} 2x - 1 = 0 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{t = cos{\mkern 1mu} 2x = 1 \Leftrightarrow x = k\pi }\\
{t = cos{\mkern 1mu} 2x = - 1 \Leftrightarrow x = k\frac{\pi }{2}}
\end{array} \Rightarrow S = \left\{ {k\pi ,k \in Z} \right\}} \right.\]
