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Well, I got this equation [itex]f(x)=\frac{2\sin{2x}}{x}[/itex] [itex] [-\pi,\pi][/itex]

So I took the 1st derivative, [itex]f'(x)=\frac{2(2x\cos{2x} - \sin{2x})}{x^2}[/itex]

Then I set that equal to 0, and got [itex]0=2x\cos{2x} - \sin{2x}[/itex]

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.

edit: Can an mod please remove this post, seems I posted in the wrong section. :(

So I took the 1st derivative, [itex]f'(x)=\frac{2(2x\cos{2x} - \sin{2x})}{x^2}[/itex]

Then I set that equal to 0, and got [itex]0=2x\cos{2x} - \sin{2x}[/itex]

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.

edit: Can an mod please remove this post, seems I posted in the wrong section. :(

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