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Trigonometric Equation

  1. Mar 20, 2014 #1
    1. The problem statement, all variables and given/known data

    4sin2(2x) -1 = 0 Solve over 0 --> 2pi

    The attempt at a solution

    (2sin(2x) + 1) (2sin(2x) - 1) = 0

    2sin(2x) +1 = 0
    sin(2x) = -1/2
    2x = -pi/6
    x = -pi/12 and -11pi/12 which = 23pi/12 and 13pi/12.

    2sin(2x) - 1 = 0
    sin(2x) = 1/2
    2x = pi/6
    x= pi/12 and 11pi/12

    I have found these four solutions, however, I do not know how to get the other four solutions.
     
  2. jcsd
  3. Mar 20, 2014 #2
    4sin2(2x) -1 = 0

    2(2sin2(2x)-1)+1=0

    -2cos(4x)+1=0

    cos(4x) = 1/2
     
  4. Mar 20, 2014 #3

    LCKurtz

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    You have ##2x = \frac \pi 6 + 2n\pi## but also ##2x = \frac {5\pi} 6 + 2n\pi## so ##x=\frac \pi {12} + n\pi## and ##x=\frac {5\pi} {12} + n\pi##. Similarly for the other one.
     
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