Sin5x+sinx=5sin3x решите уравнение

$\displaystyle \sin5x+\sin x=5\sin3x\quad \quad \quad \quad \bigg(\sin a+\sin b=2\sin\frac{a+b}2\cos\frac{a-b}2\bigg)\\\\2\sin\frac{5x+x}2\cos\frac{5x-x}2=5\sin3x\\\\2\sin3x\cos2x=5\sin3x\\\\2\sin3x\cos2x-5\sin3x=0\\\\\sin3x(2\cos2x-5)=0\\\\\\ \left[\begin{array}{ccc}sin3x=0\\2cos2x-5=0\end{array}\right =\ \textgreater \ \left[\begin{array}{ccc}3x=\pi n; \,\,n\in Z\\cos2x=2,5\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}\displaystyle x=\frac{\pi n}3;\,\,\in Z\\\varnothing\end{array}\right$

Ответ: $\boxed{\displaystyle x=\frac{\pi n}3;\,\,\in Z}$