√3.sin2x-cos2x=2sinx

Đáp án:

\(\left\{\begin{array}{l} x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{7\pi }{18}  + \dfrac{k2\pi}{3} \end{array} \right.\) $(k\in\mathbb Z)$

Giải thích các bước giải:

\(\begin{array}{l} \sqrt 3 \sin 2x - \cos 2x = 2\sin x\\ \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 2x - \dfrac{1}{2}\cos 2x = \sin x\\ \Leftrightarrow \sin 2x.\cos \dfrac{\pi }{6} - \cos 2x.\sin \dfrac{\pi }{6} = \sin x\\ \Leftrightarrow \sin \left( {2x - \dfrac{\pi }{6}} \right) = \sin x\\ \Leftrightarrow \left[ \begin{array}{l} 2x - \dfrac{\pi }{6} = x + k2\pi \\ 2x - \dfrac{\pi }{6} = \pi - x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{7\pi }{18}  + \dfrac{k2\pi}{3} \end{array} \right. \end{array}\)

$(k\in\mathbb Z)$.