Đáp án:
1) $ x = \dfrac{\pi }{4} + k\pi $ và $x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{7} $
2) $ x = \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3}$ và $ x = \dfrac{{2\pi }}{{15}} + \dfrac{{k2\pi }}{5} $
Giải thích các bước giải:
\(\begin{array}{l} {\rm{1) }}\sin 8x - \cos 6x = \sqrt 3 \left( {\sin 6x + \cos 8x} \right)\\ \Leftrightarrow \sin 8x - \sqrt 3 \cos 8x = \sqrt 3 \sin 6x + \cos 6x\\ \Leftrightarrow \dfrac{1}{2}\sin 8x - \dfrac{{\sqrt 3 }}{2}\cos 8x = \dfrac{{\sqrt 3 }}{2}\sin 6x + \dfrac{1}{2}\cos 6x\\ \Leftrightarrow \sin \left( {8x - \dfrac{\pi }{3}} \right) = \sin \left( {6x + \dfrac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} 8x - \dfrac{\pi }{3} = 6x + \dfrac{\pi }{6} + k2\pi \\ 8x - \dfrac{\pi }{3} = \pi - 6x - \dfrac{\pi }{6} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 2x = \dfrac{\pi }{2} + k2\pi \\ 14x = \dfrac{{7\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\pi \\ x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{7} \end{array} \right.\\ 2)\,\,8\cos 2x = \dfrac{{\sqrt 3 }}{{\sin x}} + \dfrac{1}{{\cos x}} \Leftrightarrow 8\cos 2x = \dfrac{{\sqrt 3 \cos x + \sin x}}{{\sin x\cos x}}\\ \Leftrightarrow 8\cos 2x\sin x\cos x = \sqrt 3 \cos x + \sin x\\ \Leftrightarrow 4\cos 2x\sin 2x = \sqrt 3 \cos x + \sin x\\ \Leftrightarrow 2\cos 2x\sin 2x = \dfrac{{\sqrt 3 }}{2}\cos x + \dfrac{1}{2}\sin x\\ \Leftrightarrow \sin 4x = \sin \left( {\dfrac{\pi }{3} + x} \right)\\ \Leftrightarrow \left[ \begin{array}{l} 4x = x + \dfrac{\pi }{3} + k2\pi \\ 4x = \pi - x - \dfrac{\pi }{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 3x = \dfrac{\pi }{3} + k2\pi \\ 5x = \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3}\\ x = \dfrac{{2\pi }}{{15}} + \dfrac{{k2\pi }}{5} \end{array} \right. \end{array}\)
