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Thread: Trigonometric equations[1 MORE]

  1. #1
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    Trigonometric equations[1 MORE]

    Hi,

    Could someone check this for me please.

    Solve

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



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

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

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

    $\displaystyle 6a^2 - a - 2 = 0$

    $\displaystyle D = 1 - 4(6)(-2) = 49$




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

    $\displaystyle = \frac{8}{12} $

    $\displaystyle = \frac{2}{3}$

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

    $\displaystyle = \frac{-1}{2}$

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

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

    $\displaystyle 0 \le x \le {\pi}$

    $\displaystyle \Rightarrow$$\displaystyle sin{\theta} > 0$

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

    $\displaystyle {\theta} = (-1)^n $

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

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

    $\displaystyle {\theta} = {\pi} - arcsin \frac{2}{3}, n = 1$
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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

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



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

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

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

    $\displaystyle 6a^2 - a - 2 = 0$

    $\displaystyle D = 1 - 4(6)(-2) = 49$




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

    $\displaystyle = \frac{8}{12} $

    $\displaystyle = \frac{2}{3}$

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

    $\displaystyle = \frac{-1}{2}$

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

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

    $\displaystyle 0 \le x \le {\pi}$

    $\displaystyle \Rightarrow$$\displaystyle sin{\theta} > 0$

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

    $\displaystyle {\theta} = (-1)^n $

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

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

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

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

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

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

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

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

    Now the general solution,

    $\displaystyle \theta=n\pi+(-1)^{n}Sin^{-1}\frac{2}{3}$
    Last edited by Sudharaka; Feb 9th 2010 at 05:38 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, $\displaystyle Sin\theta=\frac{2}{3}$

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

    $\displaystyle \theta=arcsin\frac{2}{3}$ gives all the solutions that could be taken for $\displaystyle \theta$. When writing the general solution you must take $\displaystyle \theta $ such that it is in the first quadrant. That is why I used $\displaystyle Sin^{-1}\frac{2}{3}$. 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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