# Finding Critical Numbers

by cdhotfire
Tags: critical, numbers
 P: 193 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.
 HW Helper Sci Advisor P: 3,149 Haven't I seen this somewhere before? :)
P: 193

## Finding Critical Numbers

 Quote by Tide Haven't I seen this somewhere before? :)
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.

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