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Help Understanding Trig Equation

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Find all possible solutions:
    2cos22θ = 1 - cos2θ

    3. The attempt at a solution
    I know all my arithmetic is correct, but when it comes to giving the answer, I'm not sure how to write it.
    2cos22θ + cos2θ - 1 = 0
    (2cos2θ - 1)(cos2θ + 1) = 0
    2cos2θ = 1
    cos2θ = 1/2
    2θ = ∏/3

    cos2θ = -1
    2θ = ∏

    So θ is equal to ∏/6 and ∏/2
    But the solutions are supposed to be:
    ∏/6 + ∏n
    5∏/6 + ∏n
    ∏/2 + ∏n
    And I don't understand why. Can someone explain it to me please?
     
  2. jcsd
  3. Apr 22, 2012 #2

    SammyS

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    It's because the cosine function is periodic and is an even function.
     
  4. Apr 22, 2012 #3
    Even as in cos(-x) = cos x? What do you mean by periodic? Thanks!
     
  5. Apr 22, 2012 #4

    SammyS

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    cos(x) = cos(x + 2π) = cos(x + 4π) = cos(x + 2πk) , where k is an integer.
     
  6. Apr 23, 2012 #5
    Ohhhhhh, I see, thank you! What is sin(x) then, because I know sin is odd, but I'm not sure how that looks.
     
  7. Apr 24, 2012 #6
    An odd function means that f(-x) = -f(x). Graphically, f(-x) is just f(x) flipped across the x-axis or y-axis. Also, f(x) has rotational symmetry in that it's unchanged if you rotate it 180°.
     
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