Critical Numbers for f(x)=2sin(2x)/x on [-pi,pi]

[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.

 

[cdhotfire]

Please.

 

[Tide]

Haven't I seen this somewhere before? :)

 

[cdhotfire]

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.