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Math Help - Solve the Equation exactly over the interval [0, 2pi].

  1. #1
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    Solve the Equation exactly over the interval [0, 2pi].

    Solve the Equation exactly over the interval [0, 2pi]

    the problems i do not understand are:
    2sin^2x + sin x - 1 = 0
    2sinx cosx + cos(x) = 0
    2sin^2x - sin x = 0

    if anyone could help me i would appreciate it alot. thanks
    Last edited by mr fantastic; November 29th 2011 at 03:50 PM. Reason: Title.
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  2. #2
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    Re: Solve the Equation exactly over the interval [0, 2pi] help

    Quote Originally Posted by trigonometry424 View Post
    Solve the Equation exactly over the interval [0, 2pi]

    the problems i do not understand are:
    2sin^2x + sin x - 1 = 0
    2sinx cosx + cos(x) = 0
    2sin^2x - sin x = 0
    2sin^2x + sin x - 1 = 0

    if anyone could help me i would appreciate it alot. thanks
    What have you tried?

    1) Use the quadratic formula to get values of sin(x) and thus x.
    2) Factor out cos(x)
    3) Factor out sin(x)
    4) Same as 1

    If you're stuck post what you've tried
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    Re: Solve the Equation exactly over the interval [0, 2pi] help

    i know how to do problem 1 and 2 now i think i just need to know how to factor the 3rd one now
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  4. #4
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    e^(i*pi)'s Avatar
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    Re: Solve the Equation exactly over the interval [0, 2pi] help

    Quote Originally Posted by trigonometry424 View Post
    i know how to do problem 1 and 2 now i think i just need to know how to factor the 3rd one now
    Unless I'm missing something equation 4 is the same as equation 1 so you've solved that too.

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

    You can factor out sin(x) since it's a common factor: \sin(x)(2\sin(x)-1) = 0

    Either \sin(x) = 0 or 2\sin(x) -1 = 0

    Can you find x in those cases?
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    Re: Solve the Equation exactly over the interval [0, 2pi] help

    yea sorry it was an accident i edited my op

    and would x = 1/2?
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    Re: Solve the Equation exactly over the interval [0, 2pi] help

    Quote Originally Posted by trigonometry424 View Post
    yea sorry it was an accident i edited my op

    and would x = 1/2?
    No. \sin(x) = \dfrac{1}{2} but not x.

    It's probably easier to start with when \sin(x) = 0 and then x = 0 , \pi ,  2\pi - delete as appropriate depending on your boundary conditions

    For \sin(x) = \dfrac{1}{2} we have x = \arcsin(0.5) = \dfrac{\pi}{6} as the principal solution but there is another one to find
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  7. #7
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    Re: Solve the Equation exactly over the interval [0, 2pi] help

    ohh yea
    so the answer would be pi/6 and 5pi/6 right?
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