Đáp án:
$\left[ \begin{array}{l} x = \dfrac{\pi }{3} + k\pi \\ x = \dfrac{\pi }{6} + k2\pi \\ x = - \dfrac{\pi }{2} + k2\pi \end{array} \right.\,\,\,\,\left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l} 2{\cos ^2}x - \sqrt 3 \sin 2x + 1 = \sqrt 3 \cos x - \sin x\\ \Leftrightarrow \left( {2{{\cos }^2}x - 1} \right) - \sqrt 3 \sin 2x + 2 = \sqrt 3 \cos x - \sin x\\ \Leftrightarrow \cos 2x - \sqrt 3 \sin 2x + 2 = \sqrt 3 \cos x - \sin x\\ \Leftrightarrow \left( {\dfrac{1}{2}\cos 2x - \dfrac{{\sqrt 3 }}{2}\sin 2x} \right) + 1 = \left( {\dfrac{{\sqrt 3 }}{2}\cos x - \dfrac{1}{2}\sin x} \right)\\ \Leftrightarrow \left( {\cos 2x.\cos \dfrac{\pi }{3} - \sin 2x.\sin \dfrac{\pi }{3}} \right) + 1 = \left( {\cos x.\cos \dfrac{\pi }{6} - \sin x.\sin \dfrac{\pi }{6}} \right)\\ \Leftrightarrow \cos \left( {2x + \dfrac{\pi }{3}} \right) + 1 = \cos \left( {x + \dfrac{\pi }{6}} \right)\\ \Leftrightarrow \cos \left[ {2.\left( {x + \dfrac{\pi }{6}} \right)} \right] + 1 - \cos \left( {x + \dfrac{\pi }{6}} \right) = 0\\ \Leftrightarrow \left( {2{{\cos }^2}\left( {x + \dfrac{\pi }{6}} \right) - 1} \right) + 1 - \cos \left( {x + \dfrac{\pi }{6}} \right) = 0\\ \Leftrightarrow 2{\cos ^2}\left( {x + \dfrac{\pi }{6}} \right) - \cos \left( {x + \dfrac{\pi }{6}} \right) = 0\\ \Leftrightarrow \cos \left( {x + \dfrac{\pi }{6}} \right).\left( {2\cos \left( {x + \dfrac{\pi }{6}} \right) - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos \left( {x + \dfrac{\pi }{6}} \right) = 0\\ \cos \left( {x + \dfrac{\pi }{6}} \right) = \dfrac{1}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{6} = \dfrac{\pi }{2} + k\pi \\ x + \dfrac{\pi }{6} = \dfrac{\pi }{3} + k2\pi \\ x + \dfrac{\pi }{6} = - \dfrac{\pi }{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{3} + k\pi \\ x = \dfrac{\pi }{6} + k2\pi \\ x = - \dfrac{\pi }{2} + k2\pi \end{array} \right.\,\,\,\,\left( {k \in Z} \right) \end{array}$
