Phương trình căn bậc hai của 3 sin^2 x − cos^2 x + 1 = 0 có nghiệm là:

\[\sqrt 3 sin2x - cos2x + 1 = 0\]

\[ \Leftrightarrow \frac{{\sqrt 3 }}{2}sin2x - \frac{1}{2}cos2x + \frac{1}{2} = 0\]

\[ \Leftrightarrow sin2x.cos\frac{\pi }{6} - cos2x.sin\frac{\pi }{6} = - \frac{1}{2}\]

\[ \Leftrightarrow sin(2x - \frac{\pi }{6}) = sin( - \frac{\pi }{6})\]

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

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