# [Math] the period of $(\sin x)^3 (\sin(3x))$

functionsperiodic functionstrigonometry

If I do it in the general way of L.CM I'm getting $2\pi$ as answer
I don't know where I'm making mistake because the answer is given as $\pi$

Linearise first this trigonometric polynomial: as $\;\sin 3x=3\sin x -4\sin ^3x$, we deduce that $$\sin^3x=\frac14(3\sin x-\sin 3x),$$ whence, using the standard linearisation formulæ: \begin{align}\sin^3x\sin 3x&=\frac14(3\sin x\sin 3x-\sin^2 3x)=\frac 38(\cos 2x-\cos 4x)-\frac18(1-\cos 6x)\\ &=\frac18(\cos 6x- 3\cos 4x+3\cos 2x-1). \end{align}