Câu hỏi 2

\(\begin{array}{l}\,\,\,\,\,1 + \sin x + \cos 3x = \cos x + \sin 2x + \cos 2x\\ \Leftrightarrow \left( {1 - \cos 2x} \right) + \left( {\sin x - \sin 2x} \right) + \left( {\cos 3x - \cos x} \right) = 0\\ \Leftrightarrow 2{\sin ^2}2x + \left( {\sin x - \sin 2x} \right) - 2\sin 2x\sin x = 0\\ \Leftrightarrow 2\sin 2x\left( {\sin 2x - \sin x} \right) - \left( {\sin 2x - \sin x} \right) = 0\\ \Leftrightarrow \left( {\sin 2x - \sin x} \right)\left( {2\sin 2x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin 2x = \sin x\\\sin 2x = \dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x = x + k2\pi \\2x = \pi  - x + k2\pi \\2x = \dfrac{\pi }{6} + k2\pi \\2x = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\\x = \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{{5\pi }}{{12}} + k\pi \end{array} \right.\end{array}\)

Vậy nghiệm của phương trình là: \(x = k2\pi \), \(x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\), \(x = \dfrac{\pi }{{12}} + k\pi \), \(x = \dfrac{{5\pi }}{{12}} + k\pi \).