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Homework Help: Another trig question

  1. Oct 5, 2006 #1

    I am running into an error and I can't quite figure out what I am doing wrong.

    The question is "solve [tex]14\sin^{2}y-13=0[/tex] for y where 0<y<360.

    So I factor it down to

    [tex]14(sin(y)-\frac{\sqrt{182}}{14})(sin(y)+\frac{\sqrt{182}}{14})=0[/tex] (Thanks integral for showing me how to do this) :P

    and thus get

    [tex]y=74.5\deg or y=285.5\deg[/tex] by using arcsine, but I am missing solutions somehow. I checked my answer with my calc and it's wrong, I am missing 105.5^ and 254.5^... how though? Adding the period length of 360^ does absolutely nothing. It's probably a small detail that I am overlooking, can someone point me in the right direction? Thanks again.
  2. jcsd
  3. Oct 5, 2006 #2
    [tex] 180 - 74.5 = 105.5 [/tex]

    [tex] 360 - 105.5 = 254.5 [/tex]
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