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