I apologize. I put this up and then realized the solution. Please delete this.

Find the critical points.

$\displaystyle f(x)=x^3 - 2x - 2 \cos x$

$\displaystyle f'(x)= 3x^2 -2 + 2 \sin x$

So I set the derivatie to 0, to find the critical points, and got:

$\displaystyle 3x^2 + 2 sin x - 2 = 0$ ... looks like some kind of spooky quadratic equation.

I know I need to find how those 2 equal 0.

I don't know how, aside from lots of trial and error.

Thank yoU!

edit: solution[SIZE="7"]GRAPH!