Giải phương trình \(5\left( {\sin x + \frac{{\sin 3x + \cos 3x}}{{1 + 2\sin 2x}}} \right) = \cos 2x + 3\).
Giải chi tiết:
Điều kiện: \(\sin 2x \ne - \frac{1}{2} \Leftrightarrow \left\{ \begin{array}{l}2x \ne - \frac{\pi }{6} + k2\pi \\2x \ne \frac{{7\pi }}{6} + m2\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne - \frac{\pi }{{12}} + k\pi \\x \ne \frac{{7\pi }}{{12}} + m\pi \end{array} \right.\;\left( {k,\;m \in Z} \right)\)
\(\begin{array}{l}\;\;\;\;5\left( {\sin x + \frac{{\sin 3x + \cos 3x}}{{1 + 2\sin 2x}}} \right) = \cos 2x + 3\\ \Leftrightarrow 5\left( {\sin x + \frac{{3\sin x - 4{{\sin }^3}x + 4{{\cos }^3}x - 3\cos x}}{{1 + 2\sin 2x}}} \right) = \cos 2x + 3\\ \Leftrightarrow 5\left( {\sin x + \frac{{4\left( {\cos x - \sin x} \right)\left( {1 + \sin x\cos x} \right) + 3\left( {\sin x - \cos x} \right)}}{{1 + 2\sin 2x}}} \right) = \cos 2x + 3\\ \Leftrightarrow 5\left( {\sin x + \frac{{\left( {\cos x - \sin x} \right)\left( {1 + 4\sin x\cos x} \right)}}{{1 + 4\sin x\cos x}}} \right) = \cos 2x + 3\\ \Leftrightarrow 5\left( {\sin x + \cos x - \sin x} \right) = 2{\cos ^2}x - 1 + 3\\ \Leftrightarrow 5\cos x = 2{\cos ^2}x + 2\\ \Leftrightarrow 2{\cos ^2}x - 5\cos x + 2 = 0\\ \Leftrightarrow \left( {2\cos x - 1} \right)\left( {\cos x - 2} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x = 2\left( {VN} \right)\end{array} \right. \Leftrightarrow x = \pm \frac{\pi }{3} + k2\pi \;\;\left( {k \in Z} \right)\end{array}\)
Chọn A