Trigonometric Roots of a Polynomial After wondering on this question, I wondered how would you be able to find the roots of a polynomial, in the form $y=x^3+ax^2+bx+c$ if they are the sums of cosines?
I'm wondering if it can, too, be expressed in the form $b\cos\left(A\right)$ with $A$ larger than $2\pi$.
EDIT: I believe $y=x^3+x^2-4x+1$ to be an example.
 A: After research, I found a relatively easy method. Given the cubic $y=x^3+ax^2+bx+c$, one can find the roots by finding $p$ and $q$ where $$p=\frac {3b-a^2}{9}\tag{1}$$$$q=\frac {9ab-27c-2a^3}{54}\tag{2}$$
And plugging them into $$\cos(\theta)=\frac {q}{\sqrt{-p^3}}\tag{3}$$So the solutions are given $$\begin{cases}x_1=2\sqrt{-p}\cdot\cos\left(\frac {\theta}{3}\right)-\frac {a}{3}\\x_2=2\sqrt{-p}\cdot\cos\left(\frac {\theta+2\pi}{3}\right)-\frac {a}{3}\\x_3=2\sqrt{-p}\cdot\cos\left(\frac {\theta+4\pi}{3}\right)-\frac {a}{3}\end{cases}\tag{4}$$
With discriminant $\delta=p^3+q^2<0$.

For example: $y=x^3-x^2-9x+1$. We see that $a=-1,b=-9$ and $c=1$. So from $(1)$, $$p=\frac {3(-9)-1^2}{9}=-\frac {28}{9}\tag{5}$$ and from $(2)$, $$q=\frac {9(-1)(-9)-27+2}{54}=\frac {28}{27}\tag{6}$$
So $\cos(\theta)=\frac {1}{2\sqrt{7}}$ and the solutions are $$\begin{cases}x_1=\frac {4\sqrt{7}}{3}\cdot\cos\left(\frac {1}{3}\cdot\arccos\left(\frac {1}{2\sqrt{7}}\right)\right)+\frac {1}{3}=4\cos\left(\frac {2\pi}{7}\right)+1\\x_2=
\frac {4\sqrt{7}}{3}\cdot\cos\left(\frac {1}{3}\left(2\pi+\arccos\left(\frac {1}{2\sqrt{7}}\right)\right)\right)+\frac {1}{3}=4\cos\left(\frac {8\pi}{7}\right)+1\\x_3=\frac {4\sqrt{7}}{3}\cdot\cos\left(\frac {1}{3}\left(4\pi+\arccos\left(\frac {1}{2\sqrt{7}}\right)\right)\right)+\frac {1}{3}=4\cos\left(\frac {4\pi}{7}\right)+1\end{cases}$$
Note: The polynomial has $3$ roots because $\delta$ is less than $0$.
