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