Решите уравнение \(\sin x +\sin 2x =\cos x +\cos 2x \).

Решение

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

Ответ: \(x=\pi+2k\pi, \ \dfrac{\pi}{6} + \dfrac{2k\pi}{3}, \quad k \in \mathbb{Z}\).