Q:

Critical Points of $f(x) = \sin(3x)$ --- Help!

The problem is asking me to find the Critical Points of : 
$$f(x) = \sin(3x)$$ on the closed interval: $$[\frac{-\pi}{4},\, \frac {\pi}{3}]$$
I know that $f'= 3\cos(3x)$. the problem I seem to be having here is remembering back to pre-calculus. I'm missing some kind of rule that lets me set $f'= 3\cos(3x)$ equal to zero.
I also know that the answer turns out to be $\pm\frac{\pi}{6}$. I just can't seem to remember how to get there. If someone could please explain that to me it would be much appreciated! Thanks!

A:

Here is how: You set $3\cos(3x)=0$. All you have to do now is to solve for $x$. Solving:
$3\cos(3x)=0$ means $\cos(3x)=0$, taking the inverse cosine on both sides of the last equation we get:
$3x=\frac{\pi}{2}$ and $3x=-\frac{\pi}{2}$. (the choice of $\pi /2$ and $-\pi/2$ here is due to the given interval)
Clearly, $x=\pi /6$ and $-\pi/6$. 

