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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].

    [itex]sin6x+sin2x=0[/itex]

    2. Relevant equations

    Double Angle Formulas
    [itex]sin2x=2sinxcosx[/itex]

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

    (3 formulas for cos2x)​

    [itex]tan2x=\dfrac{2tanx}{1-tan^{2}x}[/itex]

    Sum to Product Formula
    [itex]sinA+sinB=2cos\dfrac{A+B}{2}cos\dfrac{A-B}{2}[/itex]

    3. The attempt at a solution

    [itex]sin6x+sin2x=0[/itex]

    [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]2sin4xcos2x=0[/itex]

    [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

    Mark44

    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

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Isn't that what FritoTaco did?
    From here, you could have proceeded
    sin(2x)cos2(2x)=0
    sin(2x)cos(2x)=0
    Cancelling out one cos(2x) factor is ok because we have one left, in case that is the zero factor
    sin(4x)=0
    etc.
     
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