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Math Help - How should I solve this Trigonometric equation?

  1. #1
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    Hellevoetsluis
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    Post How should I solve this Trigonometric equation?

    For math homework I have to solve the following equation. I know what the answer is, I just don't quite understand how to get there. Can you please help me?
    So the equation is:

    sin(2x - ⅓π) = -cos(x + ⅓π)

    This is how I've solved it so far:

    sin(2x - ⅓π) = -cos(x + ⅓π)
    sin(2x - ⅓π) = cos(x + 4/3π)
    sin(2x - ⅓π) = sin(x + 11/6π)

    then using sin(A) = sin(B) gives A = B + k2π and A = π - B + K2π, I got:

    2x - ⅓π = x + 11/6π + K2π and 2x - ⅓π = π - (x+11/6π) + k2π
    x = 13/6π + k2π and 3x - ⅓π = -π + k2π
    x = 13/6π + k2π and x = - 1/6π + k . ⅔π

    This I all understand. Now, however, we have to solve this on an interval of [0, 2π]

    The answer: x = 1/6π and x = π and x = 7/6π and x = 11/6π

    I don't understand how they got there. Could anyone please explain this to me in as much detail as possible?
    Thank you so much!
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  2. #2
    Super Member
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    Re: How should I solve this Trigonometric equation?

    You have chosen, which to me is, a strange method of solution, but the results look OK.
    Simply substitute any integer values for k that will give you a value for x in the required range.
    That's k=-1 for your first solution and k=1,2,3 for the second.
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