Phương trình \(\sqrt 3 {\rm{sin}}2x - {\rm{cos}}2x + 1 = 0\) có nghiệm là:

Đáp án đúng là: D

\(\sqrt 3 {\rm{sin}}2x - {\rm{cos}}2x + 1 = 0\)

\( \Leftrightarrow \frac{{\sqrt 3 }}{2}{\rm{sin}}2x - \frac{1}{2}{\rm{cos}}2x + \frac{1}{2} = 0\)

\( \Leftrightarrow {\rm{sin}}2x \cdot {\rm{cos}}\frac{\pi }{6} - {\rm{cos}}2x \cdot {\rm{sin}}\frac{\pi }{6} = - \frac{1}{2}\)

\( \Leftrightarrow {\rm{sin}}\left( {2x - \frac{\pi }{6}} \right) = {\rm{sin}}\left( { - \frac{\pi }{6}} \right)\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x - \frac{\pi }{6} = - \frac{\pi }{6} + k2\pi }\\{2x - \frac{\pi }{6} = \frac{{7\pi }}{6} + k2\pi }\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x = k2\pi }\\{2x = \frac{{4\pi }}{3} + k2\pi }\end{array} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = k\pi }\\{x = \frac{{2\pi }}{3} + k\pi }\end{array}{\rm{\;}}\left( {k \in \mathbb{Z}} \right)} \right.} \right.\)