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...