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.
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)
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.
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