Câu hỏi 8

 \(\begin{array}{l}\,\,\,\,\,\,\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x\\ \Leftrightarrow \left( {\sin x + \sin 3x} \right) + \sin 2x = \left( {\cos x + \cos 3x} \right) + \cos 2x\\ \Leftrightarrow 2\sin 2x\cos x + \sin 2x = 2\cos 2x\cos x + \cos 2x\\ \Leftrightarrow \sin 2x\left( {2\cos x + 1} \right) = \cos 2x\left( {2\cos x + 1} \right)\\ \Leftrightarrow \left( {2\cos x + 1} \right)\left( {\sin 2x - \cos 2x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}2\cos x + 1 = 0\\\sin 2x - \cos 2x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x =  - \dfrac{1}{2}\\\sin 2x = \cos 2x\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  \pm \dfrac{{2\pi }}{3} + k2\pi \\\tan 2x = 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  \pm \dfrac{{2\pi }}{3} + k2\pi \\2x = \dfrac{\pi }{4} + k\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  \pm \dfrac{{2\pi }}{3} + k2\pi \\x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy nghiệm của phương trình là: \(x =  \pm \dfrac{{2\pi }}{3} + k2\pi \), \(x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\).

Chọn A.