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Another easy?trig question

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the following equations of inequalities in the interval [0, 2pi)

    cosx + 1 = sinx


    2. Relevant equations



    3. The attempt at a solution

    cos2x + 2cosx + 1 = sin2x
    cos2x + 2cosx + 1 = 1 - cos2x
    2cos2x + 2cosx = 0
    (2cosx)(cosx + 1) = 0
    2cosx = 0 or cosx + 1 = 0
    2cosx = 0
    cosx = 0
    x = pi/2, 3pi/2

    cosx + 1 = 0
    cosx = -1
    x = pi

    Did I miss anything?
     
  2. jcsd
  3. Sep 30, 2009 #2
    You forgot to check your answers. When squaring an equation, you create the possibility of extraneous solutions appearing. In the case of your problem, 3pi/2 does not work in the original equation and should be left out of the solution set.
     
  4. Sep 30, 2009 #3
    yeh you can't be too careful squaring both sides as you risk getting extra solutions, perhaps a 'cleaner' way to do it is simply to rearrange giving;

    sinx - cosx = 1

    then proceed to right it in the form Rsin(A-B)= 1

    but your method works, like mrko said just remember to check
     
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