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Tricky Trigonometric Equation

  1. Feb 11, 2012 #1
    1. The problem statement, all variables and given/known data
    cos 2x-cos^2 x=0

    3. The attempt at a solution
    I have no idea.
     
  2. jcsd
  3. Feb 11, 2012 #2
    Try getting everything in terms of ##cos^{2}{x}##

    What is ##cos(2x)## equaled to? Hint. Double angle identity.
     
  4. Feb 11, 2012 #3
    cos2x=2cos^2 -1
    Therefore 2cos^2 x -1 -cos^2 x=0
    How to continue?
     
  5. Feb 11, 2012 #4
    Don't substitute it in yet.

    ##cos(2x)## has multiple identities; it also equals:

    $$cos(2x) = cos^{2}(x) - sin^{2} (x)$$

    How can you change the ##sin^{2} (x)## into cosines?
     
    Last edited: Feb 11, 2012
  6. Feb 12, 2012 #5
    Anyone?
     
  7. Feb 12, 2012 #6

    I like Serena

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    Homework Helper

    You already had 2cos^2 x -1 -cos^2 x=0.

    Can you replace each occurrence of (cos^2 x) by y?
    And then solve for y?
     
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