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Math Help - solving trig equation

  1. #1
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    solving trig equation

    Solve for x:

    sin(2x) = sqrt(3) * sin(x)


    Thanks.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mr_Green View Post
    Solve for x:

    sin(2x) = sqrt(3) * sin(x)


    Thanks.
    Use the trig identity sin(2x) = sinxcosx + sinxcosx = 2sinxcosx

    so sin(2x) = sqrt(3) * sin(x)
    => 2sinxcosx = sqrt(3)sinx
    => (2cosx)sinx = (sqrt(3))sinx .........i rewrote this so you would see the connection.

    To maintain equality, 2cosx must be equal to sqrt(3), so the solution to our equation is the solution to 2cosx = sqrt(3)

    => cosx = sqrt(3)/2
    => x = pi/6
    Last edited by Jhevon; March 3rd 2007 at 11:47 AM.
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  3. #3
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    to signify all answers would it be pi/6 + 2pi*K

    thanks
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mr_Green View Post
    to signify all answers would it be pi/6 + 2pi*K

    thanks
    yea
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    Use the trig identity sin(2x) = sinxcosx + sinxcosx = 2sinxcosx

    so sin(2x) = sqrt(3) * sin(x)
    => 2sinxcosx = sqrt(3)sinx
    => (2cosx)sinx = (sqrt(3))sinx .........i rewrote this so you would see the connection.

    To maintain equality, 2cosx must be equal to sqrt(3), so the solution to our equation is the solution to 2cosx = sqrt(3)

    => cosx = sqrt(3)/2
    => x = pi/6
    You made an error what about x=0?

    Instead when you have,
    (2cosx)sinx = (sqrt(3))sinx
    You cannot divide by "sin x" which is what you implicity did.

    Rather subtract to get,
    (2cos x)sin x - (sqrt(3))sin x=0
    Factor,
    (sin x)(2cos x-sqrt(3))=0
    Thus,
    sin x=0
    2cos x-sqrt(3)=0
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