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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- Dec 8th 2008, 06:23 AMcalvin_coolidgeFind 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. - Dec 8th 2008, 06:37 AMsamer_guirguis_2000
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) - Dec 8th 2008, 06:37 AMChop Suey
Let

EDIT: Ah, Samer beat me to it. And for some reason, I can't delete my post. O_O - Dec 8th 2008, 06:41 AMmathaddict
- Dec 9th 2008, 06:56 AMcalvin_coolidge
Thank you all for your assistance!

~CC - Dec 9th 2008, 03:07 PMCrazyheaven
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. - Dec 10th 2008, 04:07 AMsamer_guirguis_2000
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