# Finding Critical Numbers

1. Nov 8, 2005

### cdhotfire

Well, I got this equation $f(x)=\frac{2\sin{2x}}{x}$ $[-\pi,\pi]$
So I took the 1st derivative, $f'(x)=\frac{2(2x\cos{2x} - \sin{2x})}{x^2}$
Then I set that equal to 0, and got $0=2x\cos{2x} - \sin{2x}$
But I do not see how to get the critical numbers, I also tried to do double angle, but that just resulted in more pain.
Any help, would be appreciated.

2. Nov 8, 2005

3. Nov 9, 2005

### Tide

Haven't I seen this somewhere before? :)

4. Nov 9, 2005

### cdhotfire

ya, i know, i though no one visited the mathematics section, because i was only able to join it by using the search tool. Also, i put a note for someone to delete my post in the other section.

btw, i posted back on you reply.:tongue2: