How to find the maximum value of $12\sin x -9\sin^2x$ How to find the maximum value of $12\sin x -9\sin^2x$ ; 
My approach : 
This can be written as $-[(3\sin x -2)^2-4]$.
It means that the function will be maximum when $(3\sin x-2)^2 <4$ due to negative sign outside bracket. 
But I am not getting how to proceed from here, please suggest. Thanks.

Note
$$ 12\sin x-9\sin^2x=3\sin x(4-3\sin x)\le \left(\frac{3\sin x+(4-3\sin x)}{2}\right)^2=4 $$
and the equal sign holds if and only if $3\sin x=4-3\sin x$ or $\sin x=\frac{2}{3}$.
Thus the max is 4 when $\sin x=\frac{2}{3}$.
