Question

Find the general solution of the following equation:

\[\sin 9x = \sin x\]

Answer 1

We have:

\[\sin9x = \sin x\]

⇒ \[\sin9x - \sin x = 0\]

⇒ \[2 \sin \left( \frac{9x - x}{2} \right) \cos \left( \frac{9x + x}{2} \right) = 0\]

⇒ \[\sin \frac{8x}{2} = 0\] or \[\cos \frac{10x}{2} = 0\]

⇒ \[\sin 4x = 0\] or \[\cos 5x = 0\]

⇒ \[4x = n\pi\]

\[n \in Z\] or \[5x = (2n + 1)\frac{\pi}{2}\],

\[n \in Z\]

⇒ \[x = \frac{n\pi}{4}\],

\[n \in Z\] or \[x = (2n + 1)\frac{\pi}{10}\],

\[n \in Z\]