Đáp án: B
Giải thích các bước giải:
$\begin{array}{l}
\cos 2x + \sqrt 3 \sin 2x + \sqrt 3 \sin x - \cos x = 2\\
\Rightarrow \dfrac{1}{2}\cos 2x + \dfrac{{\sqrt 3 }}{2}\sin 2x + \dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x = 1\\
\Rightarrow \cos \left( {2x - \dfrac{\pi }{3}} \right) + \sin \left( {x - \dfrac{\pi }{6}} \right) = 1\\
\Rightarrow \cos \left[ {2\left( {x - \dfrac{\pi }{6}} \right)} \right] + \sin \left( {x - \dfrac{\pi }{6}} \right) = 1\\
\Rightarrow 1 - 2{\sin ^2}\left( {x - \dfrac{\pi }{6}} \right) + \sin \left( {x - \dfrac{\pi }{6}} \right) = 1\\
\Rightarrow \sin \left( {x - \dfrac{\pi }{6}} \right).\left( {1 - 2\sin \left( {x - \dfrac{\pi }{6}} \right)} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin \left( {x - \dfrac{\pi }{6}} \right) = 0\\
\sin \left( {x - \dfrac{\pi }{6}} \right) = \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{6} = k\pi \\
x - \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{3} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\\
\Rightarrow {x_{\min }}\left( {x > 0} \right):x = \dfrac{\pi }{6}
\end{array}$
