√2(sinx-2cosx)=2-sinx

\(\begin{array}{l} \sqrt 2 \left( {\sin x - 2\cos x} \right) = 2 - \sin x\\ \Leftrightarrow \sqrt 2 \sin x + \sin x - 2\sqrt 2 \cos x = 2\\ \Leftrightarrow \left( {\sqrt 2 + 1} \right)\sin x - 2\sqrt 2 \cos x = 2\\ \Leftrightarrow \frac{{\sqrt 2 + 1}}{{\sqrt {11 + 2\sqrt 2 } }}\sin x - \frac{{2\sqrt 2 }}{{\sqrt {11 + 2\sqrt 2 } }}\cos x = \frac{2}{{\sqrt {11 + 2\sqrt 2 } }}\\ \Leftrightarrow \sin (x - \alpha ) = \frac{2}{{\sqrt {11 + 2\sqrt 2 } }}\\ \Leftrightarrow \left[ \begin{array}{l} x - \alpha = \arcsin \frac{2}{{\sqrt {11 + 2\sqrt 2 } }} + k2\pi \\ x - \alpha = \pi - \arcsin \frac{2}{{\sqrt {11 + 2\sqrt 2 } }} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \alpha + \arcsin \frac{2}{{\sqrt {11 + 2\sqrt 2 } }} + k2\pi \\ x = \alpha + \pi - \arcsin \frac{2}{{\sqrt {11 + 2\sqrt 2 } }} + k2\pi \end{array} \right. \end{array}\)