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Math Help - Solving Trigonometric Equation

  1. #1
    Senior Member vaironxxrd's Avatar
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    Solving Trigonometric Equation

    Hello, again, Everyone.

    I have the following problem

    2sin^2x = 2 + cos x

    My attempt:

    2sin^2x = 2 + cos x

     2sin^2x - cos x - 2 = 0

    x = \frac{3pi}{6}
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  2. #2
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    Re: Solving Trigonometric Equation

    Hello! ^^

    2\sin^2x-\cos x-2=0\Rightarrow 2(1-\cos^2x)-\cos x-2=0

    \Rightarrow 2-2\cos^2x-\cos x-2=0\Rightarrow 2\cos^2x+\cos x=0

    \Rightarrow \cos x(2\cos x+1)=0 \Rightarrow

    1. \cos x=0\Rightarrow x=\pm \frac{\pi}{2}+2k\pi, k \in \mathbb{Z}

    2. \cos x=-\frac{1}{2}\Rightarrow x = \pm \frac{2\pi}{3}+2k\pi, k \in \mathbb{Z}

    \Rightarrow x \in \left \{ \pm \frac{\pi}{2}+2k\pi \, |\, k \in \mathbb{Z}  \right \}\cup \left \{  \pm \frac{2\pi}{3}+2k\pi \, |\, k \in \mathbb{Z}  \right \}


    Those solutions are for x \in \mathbb{R}. If x \in [0, 2\pi), then the solutions are \frac{\pi}{2}, \frac{3\pi}{2}, \frac{2\pi}{3}, \frac{4\pi}{3}.
    Last edited by veileen; February 20th 2013 at 10:47 AM.
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  3. #3
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    Re: Solving Trigonometric Equation

    Vaironxxrd

    transfer the 2 to the left side and then substitute -2 with the basic trigonometric identity (cosx)^2+(sinx)^2=1 to find 2(cosx)^2+cosx=0
    then factorise to get cosx[2cosx+1]=0..then solve it it is easy..try....it
    MINOAS
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