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Math Help - Solving trignometric equation

  1. #1
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    Solving trignometric equation

    Hi, the problem is:

    solve on the interval 0 ≤ x ≤ 2pi

    2 sin^2 x - 3 sin x -2 = 0

    I factored the equation to (2x + 1)(x - 2)

    then separated to 2x + 1 = 0 and x - 2 = 0

    finally sin x = -1/2 and sin x = 2

    I figured out one of the x's equals 210 degrees or 7pi/6. Sin x = 2 does not seem like a solution. The given choices each have more than one possible solution, and I have only found one. How do I find the other ones?
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  2. #2
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    Re: Solving trignometric equation

    review your unit circle ...

    \sin{x} = -\frac{1}{2} at x=\frac{7\pi}{6} and x=\frac{11\pi}{6}
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  3. #3
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    Re: Solving trignometric equation

    You are right to discard Sin x = 2. But there are two solutions for Sin x = -1/2.
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  4. #4
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    Re: Solving trignometric equation

    Quote Originally Posted by Frederick View Post
    Hi, the problem is:

    solve on the interval 0 ≤ x ≤ 2pi

    2 sin^2 x - 3 sin x -2 = 0

    I factored the equation to (2x + 1)(x - 2)

    then separated to 2x + 1 = 0 and x - 2 = 0
    A technical point- since x appears in the orginal equation, it is not a good idea to use "x" to represent "sin(x)". Use some other letter.

    finally sin x = -1/2 and sin x = 2

    I figured out one of the x's equals 210 degrees or 7pi/6. Sin x = 2 does not seem like a solution. The given choices each have more than one possible solution, and I have only found one. How do I find the other ones?
    Follow Math Help Forum on Facebook and Google+

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