Решите уравнение \(\sin 7x + \sin 6x = \sin x\).

Решение

\[\sin 7x +\sin 6x = \sin x \Leftrightarrow 2\sin\dfrac{13x}{2}\cos\dfrac x2 = 2\sin\dfrac x2 \cos\dfrac x2
\Leftrightarrow \]
\[ \Leftrightarrow
\cos \dfrac x2 \cdot \left(\sin\dfrac{13x}{2}\ -\sin \dfrac x2\right)=0
\Leftrightarrow
\left[ \begin{array}{l}
\dfrac x2 = \dfrac {\pi}{2}+k\pi\\
\dfrac{13x}{2}=\dfrac x2 +2k\pi \\
\dfrac{13x}{2}=\ - \dfrac x2 + (2k+1)\pi
\end{array} \right. , \quad k \in \mathbb Z
\Leftrightarrow \]
\[ \Leftrightarrow
\left[ \begin{array}{l}
x = (2k+1)\pi\\
6x=2k\pi \\
7x=(2k+1)\pi
\end{array} \right. , \quad k \in \mathbb Z
\Leftrightarrow
\left[ \begin{array}{l}
x = \dfrac{k\pi}{3}\\
x=\dfrac{(2k+1)\pi}{7}
\end{array} \right. , \quad k \in \mathbb Z.\]

Ответ: \(\dfrac{\pi k}{3}, \ \dfrac{(2k + 1)\pi}{7}, \quad k \in \mathbb{Z}\).