\({\sin ^2}x - \left( {1 + \sqrt 3 } \right)\sin x\cos x + \sqrt 2 {\cos ^2}x = 0\)
Nhận thấy \(\cos x = 0\) không thỏa mãn phương trình nên ta có:
\(\begin{array}{l}
\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - \left( {1 + \sqrt 3 } \right)\frac{{\sin x\cos x}}{{{{\cos }^2}x}} + \sqrt 2 \frac{{{{\cos }^2}x}}{{{{\cos }^2}x}} = 0\\
\Rightarrow {\tan ^2}x - \left( {1 + \sqrt 3 } \right)\tan x + \sqrt 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = \frac{{1 + \sqrt 3 + \sqrt {4 + 2\sqrt 3 - 4\sqrt 2 } }}{2}\\
\tan x = \frac{{1 + \sqrt 3 - \sqrt {4 + 2\sqrt 3 - 4\sqrt 2 } }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \arctan \left( {\frac{{1 + \sqrt 3 + \sqrt {4 + 2\sqrt 3 - 4\sqrt 2 } }}{2}} \right) + k\pi \\
x = \arctan \left( {\frac{{1 + \sqrt 3 - \sqrt {4 + 2\sqrt 3 - 4\sqrt 2 } }}{2}} \right) + k\pi
\end{array} \right.\\
\end{array}\)
