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Math Help - trigonometric equations

  1. #1
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    Question trigonometric equations

    1.) 3 sinx = sin2x cosx

    3 ( 1-cosx) = 2sinx cosx cosx


    2.) tgx -1/tgx - 2/√3 = 0
    y - 1/y - 2/√3 = 0



    Hi, I try to solve these equations but Idon't know what I should do next. Can you help me? Thanks(Wink)
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  2. #2
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    Re: trigonometric equations

    It would help if you told us what you tried and where you got stuck.

    (Substituting y for tan(x) in (2) was a good idea. Now multiply both sides of the equation by y.)
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  3. #3
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    Re: trigonometric equations

    Quote Originally Posted by Atalante View Post
    1.) 3 sinx = sin2x cosx

    3 ( 1-cosx) = 2sinx cosx cosx


    2.) tgx -1/tgx - 2/√3 = 0
    y - 1/y - 2/√3 = 0



    Hi, I try to solve these equations but Idon't know what I should do next. Can you help me? Thanks
    1.
    \displaystyle \begin{align*} 3\sin^2{(x)} &= \sin{(2x)}\cos{(x)} \\ 3\sin^2{(x)} &= 2\sin{(x)}\cos{(x)}\cos{(x)} \\ 3\sin^2{(x)} &= 2\sin{(x)}\cos^2{(x)} \\ 3\sin^2{(x)} &= 2\sin{(x)} \left[ 1 - \sin^2{(x)} \right] \\ 3\sin^2{(x)} &= 2\sin{(x)} - 2\sin^3{(x)} \\ 2\sin^3{(x)} + 3\sin^2{(x)} - 2\sin{(x)} &= 0 \\ \end{align*}

    This can now be solved by factorising.
    Thanks from puresoul
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  4. #4
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    Re: trigonometric equations

    1.) I want to use a substitution but I can't because of 2 sinx. I need cosx there. cosx I can write like 1-sinx but there is only cosx. I must chang and I must change cosx to sinx but I don't know how.
    2.) y - 1 - 2/√3y = 0
    y - 2/√3y - 1 = 0

    D= b- 4ac
    D= (-2/√3)-4 * 1 * (-1)
    √D = 4 √3/3
    x1,2 = -b √D/2a
    x1,2 = 2/ √3 4 √3/3:2
    x1 = √3....................tg is π/3
    x2= - √3/3..............tg is 5/6π and 11/6π. Is that solition right?
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  5. #5
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    Re: trigonometric equations

    Quote Originally Posted by Atalante View Post
    1.) I want to use a substitution but I can't because of 2 sinx. I need cosx there. cosx I can write like 1-sinx but there is only cosx. I must chang and I must change cosx to sinx but I don't know how.
    Did you not bother to read my post directly above this one?
    Thanks from Atalante
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  6. #6
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    Re: trigonometric equations

    It's ok now:-) The post came while I was writing. Thank you very much.
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