A. \(x = k2\pi ;\,\,x = \dfrac{{2\pi }}{3} + k2\pi \)
B. \(x = - \dfrac{\pi }{3} + k2\pi ;\,\,x = \dfrac{{4\pi }}{3} + k2\pi \)
C. \(x = - \dfrac{\pi }{6} + k2\pi ;\,\,x = \dfrac{\pi }{2} + k2\pi \)
D. \(x = \dfrac{\pi }{6} + k2\pi ;\,\,x = \dfrac{{5\pi }}{6} + k2\pi \)
A
Ta có: \(\sqrt 3 \sin x + \cos x = 1\) \( \Leftrightarrow 2\left( {\frac{{\sqrt 3 }}{2}\sin x + \frac{1}{2}\cos x} \right) = 1\) \( \Leftrightarrow 2\sin \left( {x + \dfrac{\pi }{6}} \right) = 1\)
\( \Leftrightarrow \sin \left( {x + \dfrac{\pi }{6}} \right) = \dfrac{1}{2}\) \( \Leftrightarrow \left[ \begin{array}{l}x + \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\x + \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right. \) \(\Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\;\left( {k \in \mathbb{Z}} \right)\)
Chọn đáp án A.
Câu hỏi trên thuộc đề trắc nghiệm dưới đây !
Copyright © 2021 HOCTAP247