Đáp án:
\(\left\{ \begin{array}{l} x = - \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{7\pi }}{{18}} + k2\pi \end{array} \right.(k \in \mathbb Z)\)
Lời giải:
\(\begin{array}{l} \sqrt 3 \cos \left( {x - \dfrac{\pi }{2}} \right) + \sin \left( {x - \dfrac{\pi }{2}} \right) = 2\sin 2x\\ \Rightarrow \sqrt 3 \left( {\cos x\cos \dfrac{\pi }{2} + \sin x\sin \dfrac{\pi }{2}} \right) + \left( {\sin x\cos \dfrac{\pi }{2} - \cos x\sin \frac{\pi }{2}} \right) \\= 2\sin 2x\\ \Rightarrow \sqrt 3 \sin x - \cos x = 2\sin 2x\\ \Rightarrow \dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x = \sin 2x\\ \Rightarrow \sin \left( {x - \dfrac{\pi }{6}} \right) = \sin 2x\\ \Rightarrow \left[ \begin{array}{l} x - \dfrac{\pi }{6} = 2x + k2\pi \\ x - \dfrac{\pi }{6} = \pi - 2x + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = - \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{7\pi }}{{18}} + k2\pi \end{array} \right.(k \in \mathbb Z)\\ \end{array}\)
