giúp mình giải bài: sin10x - cos6x = căn 3(sin6x - cos10x)

$\begin{array}{l} \sin 10x - \cos 6x = \sqrt 3 \left( {\sin 6x - \cos 10x} \right)\\ \Leftrightarrow \sin 10x + \sqrt 3 \cos 10x = \sqrt 3 \sin 6x + \cos 6x\\ \Leftrightarrow \frac{1}{2}\sin 10x + \frac{{\sqrt 3 }}{2}\cos 10x = \frac{{\sqrt 3 }}{2}\sin 6x + \frac{1}{2}\cos 6x\\ \Leftrightarrow \sin \left( {10x + \frac{\pi }{3}} \right) = \sin \left( {6x + \frac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} 10x + \frac{\pi }{3} = 6x + \frac{\pi }{6} + k2\pi \\ 10x + \frac{\pi }{3} = \pi - 6x - \frac{\pi }{6} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 4x = - \frac{\pi }{6} + k2\pi \\ 16x = \frac{\pi }{2} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \frac{\pi }{{24}} + \frac{{k\pi }}{2}\\ x = \frac{\pi }{{32}} + \frac{{k\pi }}{8} \end{array} \right. \end{array}$