Đáp án: $x = \dfrac{{3\pi }}{8} + k\pi (k \in Z)$ và $x = \dfrac{{5\pi }}{{24}} + k\pi (k \in Z)$
Giải thích các bước giải:
$\begin{array}{l}
2\sqrt 3 \sin \left( {x - \dfrac{\pi }{8}} \right).\cos \left( {x - \dfrac{\pi }{8}} \right) + 2{\cos ^2}\left( {x - \dfrac{\pi }{8}} \right) = \sqrt 3 + 4\left( {{{\sin }^2}x + \cos \left( {\dfrac{\pi }{3} - x} \right)\cos \left( {\dfrac{\pi }{3} + x} \right)} \right)\\
\Leftrightarrow \sqrt 3 .\sin \left( {2x - \dfrac{\pi }{4}} \right) + 1 + \cos \left( {2x - \dfrac{\pi }{4}} \right) = \sqrt 3 + 4{\sin ^2}x + 2.\left( {\cos \left( {\dfrac{{2\pi }}{3}} \right) + \cos \left( {2x} \right)} \right)\\
\Leftrightarrow \sqrt 3 .\sin \left( {2x - \dfrac{\pi }{4}} \right) + \cos \left( {2x - \dfrac{\pi }{4}} \right) = \sqrt 3 - 1 + 4{\sin ^2}x + 2\left( {\dfrac{{ - 1}}{2} + \cos \left( {2x} \right)} \right)\\
\Leftrightarrow 2\left( {\dfrac{{\sqrt 3 }}{2}\sin \left( {2x - \dfrac{\pi }{4}} \right) + \dfrac{1}{2}\cos \left( {2x - \dfrac{\pi }{4}} \right)} \right) = \sqrt 3 - 2\left( {1 - 2{{\sin }^2}x} \right) + 2\cos \left( {2x} \right)\\
\Leftrightarrow 2.\cos \left( {2x - \dfrac{\pi }{4} - \dfrac{\pi }{3}} \right) = \sqrt 3 \\
\Leftrightarrow \cos \left( {2x - \dfrac{{7\pi }}{{12}}} \right) = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{{7\pi }}{{12}} = \dfrac{\pi }{6} + k2\pi \\
2x - \dfrac{{7\pi }}{{12}} = \dfrac{{ - \pi }}{6} + k2\pi
\end{array} \right.(k \in Z) \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{3\pi }}{8} + k\pi \\
x = \dfrac{{5\pi }}{{24}} + k\pi
\end{array} \right.(k \in Z)\\
\end{array}$
Phương trình có 2 họ nghiệm:
$x = \dfrac{{3\pi }}{8} + k\pi (k \in Z)$ và $x = \dfrac{{5\pi }}{{24}} + k\pi (k \in Z)$
