Solving an exotic trigonometric function or a sextic

Question

I'm an aerospace engineering student and I've been doing some analytical work on an interdisciplinary problem involving orbital mechanics and electromagnetism. In the final part of my work, I ended up with those four equations below

$$ e \cos\left(f - \omega \right) - 4 \cos\left(2 f + \omega \right) - 5 e \cos\left(3 f + \omega\right) =0;$$

$$ e \sin\left(f - \omega \right) + 2 e\cos\left( f+\omega \right) - 4 \sin\left(2 f + \omega \right) - 5 e \sin\left(3 f + \omega\right) =0;$$

$$-e \cos \omega + \cos\left(f + \omega \right) + 2 e \cos\left(2 f + \omega \right)=0;$$

$$-e \sin \omega + \sin\left(f + \omega \right) + 2 e \sin\left(2 f + \omega \right)=0,$$ wher $e$ is the orbital eccentricity and $\omega$ is the argument of the periapsis. $e$ is always smaller than $1$ for my application, a Keplerian elliptical orbit. $\omega$ can have any value within the domain $[0,2\,\pi]$.

The first two equations are the roots of the electric current when the satellite does not have rotation. The last two equations are the roots of the electric current for the satellite when it has a rotation equal to the orbital period i.e. the satellite always keeps pointing toward Earth's center along the orbital path. I'm using Wolfram Mathematica, and for the first two equations, the software gave me that the roots for the first and second equations can be given by the roots of the following polynomials:

$$1 - e^2 + \cos 2 \omega + e^2 \cos 2 \omega + \left(12 e + 4 e \cos 2 \omega\right) x + \left(-8 + 27 e^2 + 5 e^2 \cos 2 \omega\right) x^2 + \left(-50 e - 2 e \cos 2 \omega\right) x^3 + \left(8 - 75 e^2 - 5 e^2 \cos2 \omega \right)x^4 + 40 e x^5 + 50 e^2 x^6=0,$$

$$1 - 4 e^2 - \cos 2 \omega - 4 e^2 \cos 2 \omega + 16 e x + (-8 + 40 e^2 + 8 e^2 \cos 2 \omega) x^2 + (-54 e - 2 e \cos 2 \omega) x^3 + (8 - 85 e^2 - 5 e^2 \cos2 \omega) x^4 + 40 e x^5 + 50 e^2 x^6=0.$$ Unfortunately, it just gave me those polynomials solutions for the non-rotating satellite. I'd like to know if have any method to solve the four first trigonometric equations, or at least the 2 polynomials that apply for the non-rotating satellite only. I've been able to find some specific solutions for arbitrary $e$ with $\omega=0$ and for arbitrary $e$ with $\omega=\pi/2$, which is just a mirrored version of the first solution I mentioned, i.e. arbitrary $e$ with $\omega=0$.

If anyone knows a method for solving those equations, I would be eternally grateful, even if it depends on some non-elementary functions like the Gamma function or the Zeta function. I dedicated so hard to this work and I'd like to have 100% analytical solutions, even though my engineers and colleagues won't use them because it is probably much simpler to find those roots with numerical methods like Newton's, that's how I'm verifying the validity of the already found solutions.

I appreciate your time to read this. Thanks for your support in advance!

Answer 1

I don’t know if this helps, but you can solve for $f$ equations (3) and (4) independently of the first two equations:

You have $$2e\cos(2f+\omega)=e\cos \omega-\cos(f+\omega)$$ and $$2e\sin(2f+\omega)=e\sin\omega-\sin(f+\omega)$$

Squaring and adding these gives: $$4e^2=e^2+1-2[e\cos\omega\cos(f+\omega)+e\sin\omega\sin(f+\omega)]$$ $$\implies 3e^2=1-2e\cos f$$

Answer 2

Using $(3)$, we have $$\cos (2 \omega )=$$ $$\frac{50 e^2 x^6+\left(8-75 e^2\right) x^4+\left(27 e^2-8\right) x^2-e^2+40 e x^5-50 e x^3+12 e x+1 } { 5 e^2 x^4-5 e^2 x^2-e^2+2 e x^3-4 e x-1}$$

Using $(4)$, we have $$\cos (2 \omega )=$$ $$\frac{50 e^2 x^6+40 e x^5+\left(8-85 e^2\right) x^4-54 e x^3+\left(40 e^2-8\right) x^2+16 e x-4 e^2+1 } {5 e^2 x^4+2 e x^3-8 e^2 x^2+4 e^2+1}$$

Equating, and hoping that the denominator will not make any problem, we have $$0=\left(e^2 x^2-2 e x-2 e^2-1\right)\times $$ $$ \left(50 e^2 x^6+40 e x^5+8\left(1-10 e^2\right) x^4-52 e x^3+\left(33e^2-8\right) x^2+14 e x+\left(1-2e^2\right)\right)$$

The first factor has solutions $$x_\pm=\frac{1\pm \sqrt{2(e^2+1)}}{e}$$

The problem is not only to solve for $x$ but also to keep the solution(s) which give an acceptable result for $\cos (2 \omega )$.

Now, my questions are :

 `do you want the largest root for $x$ ?`
what is the range of $e$ ?
what is the range of $x$ ? 

Knowing the answers, I think that a few things could be done.

Edit

Beside the solutions of the quadratic, playing with a few values of $e$, what I noticed is that $$x_*=\frac{45 e^2+56 e+15}{2 \left(23 e^2+29 e+8\right)}$$ approximates the largest solution of the sextic.

Update

Let $F(x,e)$ be the sextic equation. Drawing the level curve $F(x,e)=0$ for $0\leq e\leq 1$, there are four continuous, increasing and not interescting lines which seem to be $$ +0.92 < x_1 < +0.96\qquad\qquad +0.38 < x_2 < +0.65$$

$$ -0.40 < x_3 < +0.07\qquad\qquad -0.93 < x_4 < -0.52$$

The first one is the one I almost identified.

The contour plot did not allow me to see the other but they exist. Computing the roots with a $\Delta e=0.01$ revealed that there are also two other real roots in some regions.

Some estimates

A better approximation of the largest root is $$x_*^{(1)}=\frac{1767 e^4+4336 e^3+3827 e^2+1442 e+200}{1813 e^4+4486e^3+4005 e^2+1532 e+216}$$ varies from $\frac{25}{27}$ (which is $0.926$) to $\frac{2893}{3013}$ (which is $0.960$). Since $\forall e$, $F(x_*^{(1)},e) >0$ and $F''(x_*^{(1)},e) >0$, by Darboux theorem, Newton method will converge without any overshoot during iterations. All of that means that $x_*$ is an underestimate of the solution.

For the next root
$$x_*^{(2)}=\frac{19839 e^4+28208 e^3+56432 e^2+20608 e+3584}{44618 e^4+34672e^3+80928 e^2+40448 e+9216}$$ varies from $\frac{7}{18}$ (which is $0.389$) to $\frac{128671}{209882}$ (which is $0.613$). The situation is not as good as for $x_*^{(1)}$ with respect to Darboux theorem.