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Math Help - Solving Trigonometric equations

  1. #1
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    Solving Trigonometric equations

    Find all the vlaues of x in  [0,2\pi] that are solutions of the equation:

     cos \frac{1}{2}x= \frac{-1}{2}

    Attempt:

     \frac{1}{2}x=arccos \frac{-1}{2}

     \frac{1}{2}x=120 degrees
    Since cosine is negative in the 2nd and 3rd quadrant
    2nd quadrant
    180-120=60
     \frac{1}{2}x=60
    x=120 degrees
    3rd quadrant
    No solution



    Is my attempt correct?
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  2. #2
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    Lexington, MA (USA)
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    Hello, mj.alawami!

    I understand none of your last paragraph . . .


    Find all the vlaues of x in  [0,2\pi] that are solutions of the equation:

     \cos\tfrac{x}{2}\:=\: \text{-}\tfrac{1}{2}


    Attempt:

     \frac{x}{2}\:=\:\arccos\left(\text{-}\tfrac{1}{2}\right)

     \frac{x}{2}\:=\:120^o

    Since cosine is negative in the 2nd and 3rd quadrant
    2nd quadrant
    180 - 120 = 60
     \tfrac{x}{2}\,=\,60 . . . . . . . . . . ??
    x = 120
    3rd quadrant
    No solution

    We have: . \frac{x}{2} \:=\:\arccos\left(\text{-}\tfrac{1}{2}\right)

    Then: . \frac{x}{2} \;=\;\hdots\: \text{-}240^o,\:\text{-}120^u,\:120^o,\:240^o,\:\hdots

    Hence: . x \;=\;\hdots\:\text{-}480^o, \:\text{-}240^o,\:240^o,\:480^o,\:\hdots


    Therefore: no solutions.

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