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Trig in a calculus problem

  1. Sep 11, 2006 #1
    The problem I am given is:
    [​IMG]

    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-
     
  2. jcsd
  3. Sep 11, 2006 #2

    0rthodontist

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    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).
     
  4. Sep 11, 2006 #3
    :tongue2:
    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-
     
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