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Math Help - solving for x

  1. #1
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    solving for x

    solve the following for x:

    2sin^2=1 (0 < x < 2*%pi)



    what is the solution set??

    i think there are 4 answers to this , but im not sure how to do this.....
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by samtheman17 View Post
    solve the following for x:

    2sin^2=1 (0 < x < 2*%pi)



    what is the solution set??

    i think there are 4 answers to this , but im not sure how to do this.....
    Do you mean

    2\sin^2{x}=1\text{ for }(0<x<2\pi) ?

    If so

    Solve the equation, and then restrict the infinite amount of solutions to the given interval. It's lookin like \sin{x}=\pm\frac{\sqrt{2}}{2}
    Last edited by VonNemo19; November 7th 2009 at 05:13 PM.
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  3. #3
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    okay thanks,

    but how do you solve this equation?
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by samtheman17 View Post
    okay thanks,

    but how do you solve this equation?
    2\sin^2x=1

    \sin^2x=\frac{1}{2}

    \sin{x}=\pm\frac{1}{\sqrt{2}}=\pm\frac{\sqrt{2}}{2  }

    x=\arcsin{\pm\frac{\sqrt{2}}{2}}
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