Lời giải của Tự Học 365

\(\begin{array}{l}\,\,\,\,\,\,\sqrt 3 \sin 4x - \cos 4x = \sin x - \sqrt 3 \cos x\\ \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 4x - \dfrac{1}{2}\cos 4x = \dfrac{1}{2}\sin x - \dfrac{{\sqrt 3 }}{2}\cos x\\ \Leftrightarrow \sin 4x\cos \dfrac{\pi }{6} - \cos 4x\sin \dfrac{\pi }{6} = \sin x\sin \dfrac{\pi }{6} - \cos x\cos \dfrac{\pi }{6}\\ \Leftrightarrow \sin \left( {4x - \dfrac{\pi }{6}} \right) =  - \cos \left( {x + \dfrac{\pi }{6}} \right)\\ \Leftrightarrow \sin \left( {4x - \dfrac{\pi }{6}} \right) = \sin \left( {x - \dfrac{\pi }{3}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}4x - \dfrac{\pi }{6} = x - \dfrac{\pi }{3} + k2\pi \\4x - \dfrac{\pi }{6} = \dfrac{{4\pi }}{3} - x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x =  - \dfrac{\pi }{6} + k2\pi \\5x = \dfrac{{3\pi }}{2} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\x = \dfrac{{3\pi }}{{10}} + \dfrac{{k2\pi }}{5}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Đáp án cần chọn là: a