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Homework Help: Solving a Trigonometric Equation

  1. Dec 1, 2016 #1
    1. The problem statement, all variables and given/known data
    Find all solutions of the equation in the interval [itex][0, 2\pi][/itex].


    2. Relevant equations

    Double Angle Formulas

    [itex]cos2x=cos^{2}x-sin^{2}x [/itex]
    [itex]=2cos^{2}x-1 [/itex]

    (3 formulas for cos2x)​


    Sum to Product Formula

    3. The attempt at a solution


    [itex]2sin\dfrac{6x+2x}{2}cos\dfrac{6x-2x}{2}=0[/itex] (sum to product formula)

    [itex]2sin\dfrac{8x}{2}cos\dfrac{4x}{2}[/itex] (simplify)


    [itex]2sin2xcos2x[2(cos^{2}x-1)]^{2}=0[/itex] (factor)

    It became [itex][2(cos^{2}x-1)]^{2}=0[/itex] because 2sin cancels out and the 4 reduces to 2 which is left with, [itex][2(cos^{2}x-1)]^{2}[/itex] and it's squared because the cos2x is used in the Double Angle Formula. I don't know where to go from here, any suggestions? Thanks.
  2. jcsd
  3. Dec 1, 2016 #2


    Staff: Mentor

    Mistake here...
    ... and here.
    In the first mistake, ##\cos(2x) = 2\cos^2(x) - 1##, which is different from what you wrote.
    In the second mistake, "cancelling" is not a viable option, as you lose solutions. Here's a simple example:
    x(x - 1) = 0
    "Cancel" x to get x - 1 = 0, or x = 1
    This is incorrect, or at least incomplete, as x = 0 is a solution of the original equation.
  4. Dec 1, 2016 #3
    You could try:

    2x = u or x = u/2

    Then you get:

    sin u + sin 3u = 0

    If we use "sum to product formula" you get:

    A+B = 3u + u = 4u
    A-B = 2u

    Which leads to:

    2Sin 2u . Cos u = 0

    Can you solve it from here? :-)
  5. Dec 3, 2016 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Isn't that what FritoTaco did?
    From here, you could have proceeded
    Cancelling out one cos(2x) factor is ok because we have one left, in case that is the zero factor
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