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Trig question

  1. Mar 30, 2008 #1
    find all solutions to sin^2 x = 1/4 in pi/2 <= x < 2pi (i.e in quadrants 2,3 and 4)


    i understand what i need to do but dont understand sin^2 part. will i need to apply pythagoras therom here first?



    thanks
     
  2. jcsd
  3. Mar 30, 2008 #2
    [tex]\sin ^2x=(\sin x)^2[/tex]

    You can square root both sides, and finding x should be simple then.
     
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