Phương trình $2\sin 2x - 3\sqrt 6 |\sin x + \cos x| + 8 = 0$ có nghiệm là:

Đặt $|\sin x + \cos x| = t\,\left( {t \in \left[ {0 ;\sqrt 2 } \right]} \right) \Rightarrow \sin 2x = {t^2} - 1$. Khi đó phương trình trở thành:

$2{t^2} - 3\sqrt 6 t + 6 = 0 \Leftrightarrow \left[ \begin{array}{l}t = \sqrt 6 \,(L)\\t = \dfrac{{\sqrt 6 }}{2}\,(TM)\end{array} \right. \Rightarrow \left| {\sin x + \cos x} \right| = \dfrac{{\sqrt 6 }}{2} \Leftrightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) =  \pm \dfrac{{\sqrt 6 }}{2}$

$\Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) =  \pm \dfrac{{\sqrt 3 }}{2} \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin \left( { \pm \dfrac{\pi }{3}} \right) \Leftrightarrow \left[ \begin{array}{l}x + \dfrac{\pi }{4} = \dfrac{\pi }{3} + k2\pi \\x + \dfrac{\pi }{4} = \dfrac{{2\pi }}{3} + k2\pi \\x + \dfrac{\pi }{4} =  - \dfrac{\pi }{3} + k2\pi \\x + \dfrac{\pi }{4} = \dfrac{{4\pi }}{3} + k2\pi \end{array} \right.\,\left( {k \in \mathbb{Z}} \right)$

$\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{{12}} + k2\pi \\x = \dfrac{{5\pi }}{{12}} + k2\pi \\x =  - \dfrac{{7\pi }}{{12}} + k2\pi \\x = \dfrac{{13\pi }}{{12}} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{{12}} + k2\pi \\x = \dfrac{{5\pi }}{{12}} + k2\pi \\x =  - \dfrac{{7\pi }}{{12}} + k2\pi \\x = \dfrac{{13\pi }}{{12}} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{{5\pi }}{{12}} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right).$