1+2sinxcosx=sinx+2cosx

$\begin{array}{l} 1 + 2\sin x\cos x = \sin x + 2\cos x\\  \Leftrightarrow 1 - \sin x + 2\sin x\cos x - 2\cos x = 0\\  \Leftrightarrow \left( {1 - \sin x} \right) + 2\cos x\left( {\sin x - 1} \right) = 0\\  \Leftrightarrow \left( {\sin x - 1} \right)\left( {2\cos x - 1} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} \sin x = 1\\ \cos x = \dfrac{1}{2} \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{2} + k2\pi \\ x = \dfrac{\pi }{3} + k2\pi \\ x =  - \dfrac{\pi }{3} + k2\pi  \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$