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Math Help - Trig Equation

  1. #1
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    Trig Equation

    1 + 4sin[A](3sin[A] - 4) = cosec[A](cosec[A]-4)

    Find the possible values of sin[A]


    I tried writing as a quartic equation but I couldn't factorise it and I am now totally void of ideas. Could someone point me in the right direction?
    Last edited by kinhew93; December 28th 2012 at 02:41 PM.
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  2. #2
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    Re: Trig Equation

    Quote Originally Posted by kinhew93 View Post
    1 + 4sin[A](3sin[A] - 4) = cosec[A](cosec[A]-4)
    Find the possible values of sin[A]

    I have no idea if this helps, but I would try:

    \csc(A)(\csc(A)-4)=\frac{1}{\sin^2(A)}-\frac{4}{\sin(A)}
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  3. #3
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    Re: Trig Equation

    Hello, kinhew93!

    \text{Find the possible values of }\sin\theta:

    . . . 1 + 4\sin\theta (3\sin\theta - 4) \:=\: \csc\theta(\csc\theta -4)

    We have: . . . . . 1 + 4\sin\theta(3\sin\theta - 4) \;=\;\frac{1}{\sin\theta}\left(\frac{1}{\sin\theta  } - 4\right)

    n . . . . . . . . . . . 1 + 12\sin^2\!\theta - 16\sin\theta \;=\;\frac{1-4\sin\theta}{\sin^2\!\theta}

    n . . . . . . . . \sin^2\!\theta + 12\sin^4\!\theta - 16\sin^3\!\theta \;=\;1 - 4\sin\theta

    12\sin^4\theta - 16\sin^3\theta + \sin^2\theta + 4\sin\theta - 1 \;=\;0


    Factor: . (\sin\theta-1)(2\sin\theta-1)(2\sin\theta+1)(3\sin\thetau-1) \;=\;0

    Therefore: . \sin\theta \;=\;\begin{Bmatrix} 1 \\ \pm\frac{1}{2} \\ \frac{1}{3} \end{Bmatrix}
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