# 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.