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