Đặt $t=\sin x+\cos x(-\sqrt 2\le t\le \sqrt 2)$
$\begin{array}{l} {t^2} = {\left( {\sin x + \cos x} \right)^2} = 1 + 2\sin x\cos x\\ \Leftrightarrow {t^2} - 1 = \sin 2x \end{array}$
Phương trình trở thành:
$\begin{array}{l} {t^2} - 1 - 2\sqrt 2 t - 5 = 0\\ \Leftrightarrow {t^2} - 2\sqrt 2 t - 6 = 0\\ \Leftrightarrow {t^2} - 3\sqrt 2 t + \sqrt 2 t - 6 = 0\\ \Leftrightarrow t\left( {t - 3\sqrt 2 } \right) + \sqrt 2 \left( {t - 3\sqrt 2 } \right) = 0\\ \Leftrightarrow \left( {t + \sqrt 2 } \right)\left( {t - 3\sqrt 2 } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} t = - \sqrt 2 (TM)\\ t = 3\sqrt 2 (L) \end{array} \right.\\ \Rightarrow \sin x + \cos x = - \sqrt 2 \\ \Leftrightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = - \sqrt 2 \\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = - 1\\ \Leftrightarrow x + \dfrac{\pi }{4} = - \dfrac{\pi }{2} + k2\pi \\ \Leftrightarrow x = - \dfrac{{3\pi }}{4} + k2\pi \left( {k \in \mathbb{Z}} \right) \end{array}$
