What is the period of $(\sin x)^3 (\sin(3x))$? 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 $\pi$
 A: Hint:
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}
A: Correct if wrong :
$\sin ^3 x( \sin2x \cos x +\cos 2x \sin x)=$
$(1/2)(\sin^2 x) \sin^2 2x +$
$ (\sin^4 x) \cos 2x.$
Left to do: 
Find the basic periods of the individual functions above.
(What is the basic period of $\sin^2 x$, and of $\sin^4 x$ ?)
And then?
A: Your LCM method will give you a largest possible period but it could be shorter.  You need to examine the particular functions involved.  Consider the period of $sin^2(x)$.  Is it $2\pi$?  No, it is only $\pi$.  It can be rewritten in terms of $\sin(2x)$.  Here's an even simpler example.  If $f(x)$ and $g(x)$ both have period $2\pi$ then what is the period of $f(x) + g(x)$?  How about if $f(x) = \sin(x)$ and $g(x) = - \sin(x)$? 
