# Thread: Find all solutions for....

1. ## 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.

2. 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)

3. Let $\displaystyle u = \tan{x}$

$\displaystyle 3u^3-3u^2-u+1=0$

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

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

EDIT: Ah, Samer beat me to it. And for some reason, I can't delete my post. O_O

4. Originally Posted by calvin_coolidge
3tan^3x-3tan^2x-tanx+1=0
Maybe you can try this

Let $\displaystyle y = \tan x$ , thus you will have $\displaystyle 3y^3-3y^2-y+1=0$
Then solve for y and you should be able to find the solutions .

5. Thank you all for your assistance!

~CC

6. 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.

7. 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