\(\begin{array}{l}\sin x + \sin 2x = \cos x + \cos 2x\\ \Leftrightarrow \sin 2x - \cos 2x = \cos x - \sin x\\ \Leftrightarrow \sqrt 2 \sin \left( {2x - \frac{\pi }{4}} \right) = \sqrt 2 \sin \left( {\frac{\pi }{4} - x} \right)\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4} - x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{4} = \frac{\pi }{4} - x + k2\pi \\2x - \frac{\pi }{4} = \pi  - \frac{\pi }{4} + x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + \frac{{k2\pi }}{3}\\x = \pi  + k2\pi \end{array} \right.,k \in \mathbb{Z}.\end{array}\)