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Simple Trig Question

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    -sin x = √3 * cos x
    where x is [0,2π]


    2. Relevant equations



    3. The attempt at a solution
    Would it be wrong to square both sides and then factor?
    sin2 x = 3cos2 x
    1-cos2 x = 3cos2 x
    0= 4cos2 x - 1
    0 = (2cos x -1)(2cos x + 1)
    Now I solve the factors (2 solutions for each)

    Since I've squared both sides, haven't I introduced extra solutions that are not part of the original problem? Will all 4 resulting solutions be correct? Thanks.
     
  2. jcsd
  3. Feb 15, 2012 #2
    Well, when you square root something, you will get a positive and a negative answer, which you already know. Since you defined x as [0,2π], there are no negative numbers in that interval. Therefore, you may disregard the negative answers for this interval, which should leave you with two answers.
     
  4. Feb 15, 2012 #3

    tiny-tim

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    hi kscplay! :smile:
    yes, since you will also be finding the solutions to sin x = √3 * cos x

    the same x can't be a solution to both (unless sinx = 0, of course)

    i think the only way to decide which solutions to reject is to actually check each one :wink:
     
  5. Feb 15, 2012 #4

    HallsofIvy

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    But the simplest way to solve that equation is to divide both sides by -cos(x):
    [tex]\frac{sin(x)}{cos(x)}= tan(x)= -\sqrt{3}[/tex]
     
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