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Math Help - Need help solving trig. equations.

  1. #1
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    Question Need help solving trig. equations.

    How do I solve the following for y, where 0 is less than or equal to y and y is less than 360 degrees?

    11sin^2y = 13 - 3sin^2y ?


    Thank you.
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  2. #2
    Senior Member I-Think's Avatar
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    11sin^{2}y=13-3sin^{2}y
    14sin^{2}y=13
    sin^{2}y=\frac{13}{14}
    siny=\sqrt{\frac{13}{14}}
    y=sin^{-|}\sqrt{\frac{13}{14}}
    y=74.5^o, 285.35^o

    Cheers!
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  3. #3
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    Thanks! But how did you get 285.35 degrees?
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  4. #4
    Senior Member I-Think's Avatar
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    Oops!
    That should be 105.5^o
    Sorry for that mistake, misquoted an identity

    To get the second answer, use the identity: sin\theta=sin(180-\theta)

    The identity I misquoted was: sin(360-\theta)=-sin\theta
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