I recently faced a problem with a polynomial of 6th degree, a sextic. I want an analytical solution to the problem, and I read in the last few days that Kampé de Fériet functions can solve general sextic equations. I'm an engineer, so I'm not an expert on this field and can't find recent papers on the subject, i.e. solving a sextic with Kampé de Fériet functions. I can only find papers from the 1910s or before. And I don't have enough knowledge to adventure myself with such material.
I know that Newton's method could easily find the roots I need, but I'd like an exact solution, even if no one will use it due its complexity and difficult implementation.
So I'm referring to the real experts in math which may help me with this problem. It may be important for you guys to look at the equation. But I'm editing this post on my cellphone, it will be hard to write it now, but if you want to look at the equations, I can edit it as soon as I reach my computer. I mean equations, in plural, because 2 polynomials must be solved separately, there are no common solutions to both.
As requested, I'm editing the post to express the problem. The problem involves orbital dynamics, so I'll try to make a brief resume. It is the second or third time I have used Stack Exchange, so I don't know some commands, like labeling and referencing equations, so I'll do it manually. The real equations Iwanto to solve are
$$ e \cos\left(f - \omega \right) - 4 \cos\left(2 f + \omega \right) - 5 e \cos\left(3 f + \omega\right) =0;\quad \quad(1)$$
$$ e \sin\left(f - \omega \right) + 2 e\cos\left( f+\omega \right) - 4 \sin\left(2 f + \omega \right) - 5 e \sin\left(3 f + \omega\right) =0.\quad \quad(2)$$
Equations $(1)$ and $(2)$ must be solved separately. In the equations above we have 3 parameters $e$, $\omega$, and $f$. The constant $e$ is the orbital eccentricity, it is a non-negative real number, for my application, the possible values for $e$ are $0\leq e<1$, for $e=0$, I do have solutions, so in the present moment I'm just interested in the interval $0<e<1$ which means we have an elliptical orbit. The other constant is $\omega$, also called the argument of periapsis (periapsis is the point where the object has its closer approach relative to the massive planet), this angle determines where the periapsis is pointing, and its range is $0\leq \omega \leq 2 \pi$, I do have solutions for arbitrary $e$ when $\omega=0$ or $\omega=\pi/2$. The angle $f$ is our variable, it is this value that determines where the object is along its flight. These 2 equations must be solved for $f$ independently.
If you need you can read about the orbital elements at
(https://en.wikipedia.org/wiki/Orbital_elements).
For the true anomaly information
(https://en.wikipedia.org/wiki/True_anomaly).
I'm using the software Mathematica, on it it has a built-in function named Solve[], which as the name suggests, solves an equation analytically. When I use solve on equation $(1)$, i.e. Solve[$(1)$==$0$,f], the software gives me an answer that the roots of this trigonometric function is the arccosine of the roots of the following polynomial
$$1 - e^2 + \cos 2 \omega + e^2 \cos 2 \omega + \left(12 e + 4 e \cos 2 \omega\right) x + \left(-8 + 27 e^2 + 5 e^2 \cos 2 \omega\right) x^2 + \left(-50 e - 2 e \cos 2 \omega\right) x^3 + \left(8 - 75 e^2 - 5 e^2 \cos2 \omega \right)x^4 + 40 e x^5 + 50 e^2 x^6, \quad (3)$$
for equation $(2)$, it returns another polynomial given below
$$1 - 4 e^2 - \cos 2 \omega - 4 e^2 \cos 2 \omega + 16 e x + (-8 + 40 e^2 + 8 e^2 \cos 2 \omega) x^2 + (-54 e - 2 e \cos 2 \omega) x^3 + (8 - 85 e^2 - 5 e^2 \cos2 \omega) x^4 + 40 e x^5 + 50 e^2 x^6 \quad \quad \quad (4).$$
This means, if I want the roots of equations $(1)$ and $(2)$, I must solve equations $(3)$ and $(4)$, respectively. That's why my interest in solving a sextic in a general form, is to obtain closed analytical equations for any orbit. Recently (yesterday, to be more precise) I solved another set of equations like $(1)$ and $(2)$ for another configuration using the complex exponentials. I also tried to do the same thing for equations $(1)$ and $(2)$. This time I obtained another 2 polynomials
$$5e+4x-e \exp ^{2 i \omega}x^2- ex^4+4\exp^{2 i \omega}x^5 + 5 e \exp^{2 i \omega} x^6, \quad \quad \quad (5), $$
for the complex exponential form of equation $(1)$, and
$$-5e+4x+e(2+ \exp ^{2 i \omega})x^2 - e(2 +\exp^{2 i \omega})x^4 + 4\exp^{2 i \omega}x^5 + 5 e \exp^{2 i \omega} x^6, \quad \quad \quad (5), $$
for the complex exponential form of equation $(2)$, $i=\sqrt{-1}$. The solutions to the equations $(1)$ and $(2)$ are not in the form of $arccos($roots$)$ anymore, but $-i \ln($roots$)$, but it does not matter, as long as those polynomials are solved haha.
I've been reading that Abel-Rufini Theorem, or more generally Galois theory states that it is impossible to achieve a general closed form for quintics or higher polynomials in terms of arithmetic operations of the coefficients unless the polynomial satisfies certain special conditions, however, it does not mean anything about non-elementary functions, and as Wolfram Mathworld says a general sextic could be solve using Kampé de Fériet functions.
(https://mathworld.wolfram.com/SexticEquation.html)
That's why I'm interested in the subject, especially because Mathematica does not have those functions within it, so that is my hope, because I'm pretty sure the methods to solve it by radicals, can't be applicable for arbitrary values of $e$, and $\omega$, and if there are ranges where it is applicable, I'm not an expert on the field to make such analysis, as I mentioned before.
Now the question is very long, so if you read until the end, I appreciate it!
