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Math Help - Trigonometric equations[1 MORE]

  1. #1
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    Trigonometric equations[1 MORE]

    Hi,

    Could someone check this for me please.

    Solve

    6 cos^2{\theta} + sin{\theta} = 4\, for\, 0 \le x \le {\pi}



    6(1 - sin^2{\theta}) + sin{\theta} = 4

    -6sin^2{\theta} + sin{\theta} + 2 = 0

    \text{Let} sin{\theta} = a, |a| \le 1

    6a^2 - a - 2 = 0

    D = 1 - 4(6)(-2) = 49




    a_1 = \frac{-(-1) + \sqrt49}{(2)(6)}

    = \frac{8}{12}

    = \frac{2}{3}

    a_2 = \frac{-(-1) - \sqrt49}{(2)(6)}

    = \frac{-1}{2}

    sin{\theta}_1 = \frac{2}{3}

    sin{\theta}_2 = \frac{-1}{2}

    0 \le x \le {\pi}

    \Rightarrow  sin{\theta} > 0

    \Rightarrow sin{\theta} = \frac{2}{3}

    {\theta} = (-1)^n

    arcsin \frac{2}{3} + {\pi}n

    \text{Solution:}\, {\theta}  = arcsin \frac{2}{3}, n = 0

    {\theta} = {\pi} - arcsin \frac{2}{3}, n = 1
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    Hi,

    Could someone check this for me please.

    Solve

    6 cos^2{\theta} + sin{\theta} = 4\, for\, 0 \le x \le {\pi}



    6(1 - sin^2{\theta}) + sin{\theta} = 4

    -6sin^2{\theta} + sin{\theta} + 2 = 0

    \text{Let} sin{\theta} = a, |a| \le 1

    6a^2 - a - 2 = 0

    D = 1 - 4(6)(-2) = 49




    a_1 = \frac{-(-1) + \sqrt49}{(2)(6)}

    = \frac{8}{12}

    = \frac{2}{3}

    a_2 = \frac{-(-1) - \sqrt49}{(2)(6)}

    = \frac{-1}{2}

    sin{\theta}_1 = \frac{2}{3}

    sin{\theta}_2 = \frac{-1}{2}

    0 \le x \le {\pi}

    \Rightarrow  sin{\theta} > 0

    \Rightarrow sin{\theta} = \frac{2}{3}

    {\theta} = (-1)^n

    arcsin \frac{2}{3} + {\pi}n

    \text{Solution:}\, {\theta}  = arcsin \frac{2}{3}, n = 0

    {\theta} = {\pi} - arcsin \frac{2}{3}, n = 1
    Dear Hellbent,

    It would be easier if you factorize the equation other than using the general solution of a quadratic equation.

    That is, -6sin^2{\theta} + sin{\theta} + 2 = 0

    (3sin\theta-2)(2sin\theta+1)=0

    Since,~0\leq\theta\leq\pi\Rightarrow{0\leq{Sin}\th  eta\leq{1}}

    Sin\theta=\frac{2}{3}

    Now the general solution,

    \theta=n\pi+(-1)^{n}Sin^{-1}\frac{2}{3}
    Last edited by Sudharaka; February 9th 2010 at 06:38 AM.
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  3. #3
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    Cool thanks
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  4. #4
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    Quote Originally Posted by Hellbent View Post
    Cool thanks
    Dear Hellbent,

    Sorry I did'nt help you when I first saw your thread. And that must be the reason for you to send me a privet message. That is because I just came home after a tiresome days work and I did'nt have that feeling to answer threads.

    Anyway now do you know your mistake,

    When, Sin\theta=\frac{2}{3}

    \theta=arcsin\frac{2}{3} gives all the solutions that could be taken for \theta. When writing the general solution you must take \theta such that it is in the first quadrant. That is why I used Sin^{-1}\frac{2}{3}. If this seems a bit confusing please refer,

    General Solutions of Trigonometric Functions, Maths First, Institute of Fundamental Sciences, Massey University

    Hope this will help you.
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  5. #5
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    Quote Originally Posted by Sudharaka View Post
    Dear Hellbent,

    Sorry I did'nt help you when I first saw your thread. And that must be the reason for you to send me a privet message. That is because I just came home after a tiresome days work and I did'nt have that feeling to answer threads.

    Anyway now do you know your mistake,

    When, Sin\theta=\frac{2}{3}

    \theta=arcsin\frac{2}{3} gives all the solutions that could be taken for \theta. When writing the general solution you must take \theta such that it is in the first quadrant. That is why I used Sin^{-1}\frac{2}{3}. If this seems a bit confusing please refer,

    General Solutions of Trigonometric Functions, Maths First, Institute of Fundamental Sciences, Massey University

    Hope this will help you.
    I got it
    Cool, thanks.
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