What is the Solution for Finding x in a Trigonometric Equation?

[GeoMike]

The problem I am given is:
http://www.mcschell.com/prob.gif

I determined that the MVT can be applied, I found the derivative of f(x) [f'(x) = (2cos(x) + 2cos(2x)], and now I need to determine where f'(x) = 0 (the slope of the secant line through the points (pi, f(pi)) and (2pi, f(2pi)).

The problem I'm having is in determining the values for x for this equation:
cos(x) + cos(2x) = 0

I know how to find this value with a graph/calculator, but I'm having trouble finding it analytically. I've tried applying a few trigonometric identites to the second term, but I still can't get the equation into any form that makes finding x a straightforward process for me.

It's been a while since I took trig, so some of this is fuzzy. I just need some nudging in the right direction.
Thanks,
-GeoMike-

 

[0rthodontist]

So you have 0 = cos(x) + cos(2x)
cos 2x = cos^2(x) - sin^2(x)
= cos^2(x) - (1 - cos^2(x))
= 2 cos^2(x) - 1
so, 0 = cos(x) + 2 cos^2(x) - 1. This is a quadratic in cos(x).

 

[GeoMike]

I had that written down, but totally overlooked the quadratic equation.

So, the only x value in the open interval (pi, 2pi) that makes f'(x)=0 is 5pi/3

Thanks!
-GeoMike-