Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
\sin x + \sin 2x = \cos x + \cos 2x\\
\Leftrightarrow \sin x - \cos x = \cos 2x - \sin 2x\\
\Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = \sqrt 2 \cos \left( {2x + \frac{\pi }{4}} \right)\\
\Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \cos \left( {2x + \frac{\pi }{4}} \right)\\
\Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{2} - 2x - \frac{\pi }{4}} \right)\\
\Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4} - 2x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x + \frac{\pi }{4} = \frac{\pi }{4} - 2x + k2\pi \\
x + \frac{\pi }{4} = \pi - \frac{\pi }{4} + 2x + k2\pi
\end{array} \right.\\
...
\end{array}\]
