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Trigonometry Using Trigonometry for solving a Cubic equation

Cbarker1

Active member
Jan 8, 2013
241
Dear Everyone,

I need some help for this general cubic equation.

\[ax^3+bx^2+cx+d=0\]. First divide the equation by a
\[x^3+b/ax^2+c/ax+d/a=0\]

Let x=y-b/(3a)
...

$$y^3+py+q=0$$
where $$p=c/a-(b/a)^2)/3;q=d/a-(b/a*c/a)/3+(b/a)^3/27$$
$$y=cos \theta$$
then i need to use the identity for $$cos(3\theta)= 4cos^3(\theta)-3cos(\theta) $$

and that is where need the help.
What to do next?

Thank you for your patiences
Cbarker1
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

Cbarker1

Active member
Jan 8, 2013
241
I read that article already. I am still confuse with the substitution.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I read that article already. I am still confuse with the substitution.
Did you try the substitution $y=2\sqrt{-\frac p 3}\cos\theta$?
 

Cbarker1

Active member
Jan 8, 2013
241
Why Do the substitution works?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Why Do the substitution works?
It is exactly the substitution the wiki article describes.
A great mathematician thought it up.
So yes, the substitution works.