<s>[INST] finding the maxima and minima for $f(x) = \sin\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right)$ the actual question is to show that the function$f(x) = \sin\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right)$ for all x belongs to $R$ has a minimum at $x=\left(\frac{\pi}{6}\right)$ and maxima at $x=\left(\frac{\pi}{2}\right)$ and at $x=\left(\frac{-\pi}{6}\right)$.
Actually the $f'(x)$ is coming to be $\sin\left(x-\frac{\pi}{3}\right)\cos\left(x+\frac{\pi}{3}\right) + \cos\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right)=  \sin(2x)$ $\Rightarrow$x=$\frac{\pi}{2}$.
I am not getting how to find the other 2 points $x=\frac{-\pi}{6}$ and $x=\frac{\pi}{6}$.
there is only- ONE critical point which is $\frac{\pi}{2}$ 
 [/INST] Expanding first the original formula you can see better what is happening:
$$
f(x) = \sin\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right) =
\frac14-\cos^2 x.
$$
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