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Math Help - Find all solutions for....

  1. #1
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    Find all solutions for....

    3tan^3x-3tan^2x-tanx+1=0

    I started by grouping/factoring out tanx...giving me tanx(3tan^2x-3tanx-1)+1=0...can't factor it...stuck @ the moment...

    Any thoughts or a push in the right direction would be appreciated!

    Thx,

    ~CC.
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  2. #2
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    This is a polynomial of the third degree which can be solved as follows:
    let tan(x) =y
    3y^3-3y^2-y+1=0
    dividing the equation by 3 (the coefficient of the highest power)
    y^3-y^2-y/3+1/3
    the last term (+1/3) is the product of the three roots of the equation. This means that the roots should contain the value 1.
    dividing the equation by (y-1)
    then the above equation will be
    (y-1)(3y^2-1)=0
    the second bracket can be solved by the known equation of solving quadratic equations
    the root are 1,1/root(3),-1/root(3)
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  3. #3
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    Let u = \tan{x}

    3u^3-3u^2-u+1=0

    \implies 3u^2(u-1)-(u-1)=0

    \implies (3u^2-1)(u-1)=0

    EDIT: Ah, Samer beat me to it. And for some reason, I can't delete my post. O_O
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  4. #4
    MHF Contributor
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    Quote Originally Posted by calvin_coolidge View Post
    3tan^3x-3tan^2x-tanx+1=0
    Maybe you can try this

    Let y = \tan x , thus you will have 3y^3-3y^2-y+1=0
    Then solve for y and you should be able to find the solutions .
    Last edited by mr fantastic; December 10th 2008 at 04:28 AM. Reason: Fixed the quote and latex formatting
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  5. #5
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    Thank you all for your assistance!

    ~CC
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  6. #6
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    Not to hijack the thread but could someone explain this step for me.

    "the last term (+1/3) is the product of the three roots of the equation. This means that the roots should contain the value 1.
    dividing the equation by (y-1)"

    I'm not really understanding why that is and when to use it.
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  7. #7
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    Any equation of the third degree can be written in two forms:
    1- y^3+ay^2+by+c=0
    2- (y-y1)*(y-y2)*(y-y3)=0
    where y1,y2,y3 are the roots of the equation
    if you multiply these three brackets you will find that (-y1y2y3)=c provided that the coeffecient of y^3 is 1
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