Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k\pi \\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
{\sin ^2}x - \left( {1 + \sqrt 3 } \right)\sin x.\cos x + \sqrt 3 {\cos ^2}x = 0\\
\to \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - \dfrac{{\left( {1 + \sqrt 3 } \right)\sin x.\cos x}}{{{{\cos }^2}x}} + \sqrt 3 = 0\\
\to {\tan ^2}x - \left( {1 + \sqrt 3 } \right)\tan x + \sqrt 3 = 0\\
\to \left[ \begin{array}{l}
\tan x = \sqrt 3 \\
\tan x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k\pi \\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)