Trig in a calculus problem

1. Sep 11, 2006

GeoMike

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

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-

Last edited by a moderator: May 2, 2017
2. Sep 11, 2006

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).

3. Sep 11, 2006

:tongue2: