If you have an 8th degree polynomial $f(t)$, and you know that despite 8 solutions, there are 4 solutions $x_0,x_1,x_2,x_3$ but the other 4 solutions are $2*\arctan(x0)+\pi, 2*\arctan(x1)+\pi, 2*\arctan(x2)+\pi$ and $2*\arctan(x3)+\pi$, is there a way to factor the polynomial to a 4th degree polynomial?
For example, you can expand the form:
$f(t) = (t - x_0)(t - x_1)(t - x_2)(t - x_3)(t - [2\arctan(x_0) + \pi])(t - [2\arctan(x_1) + \pi])(t - [2\arctan(x_2) + \pi])(t - [2\arctan(x_3) + \pi])$
And then equate coefficients of $f(t)$
But it seems a system of equations involving $x_0,x_1,x_2,x_3,\arctan(x_0),\arctan(x_1),\arctan(x_2),\arctan(x_3)$ may be difficult to solve unless such a technique is known?
Original problem: An equation of form $Z+A*\sin(x)+B*\cos(x)+C*\sin^2(x)+D*\sin(x)\cos(x)+E*\sin^3(x)+F*\sin^2(x)\cos(x)+G*\sin^4(x)+H*\cos(x)*\sin^3(x)$
leads to such a Weierstrass equation where the solutions will be ${x, x+\pi}$ for $x$ in the solutions. It is a general form (after $\cos^2(x)$ is replaced by $(1-\sin^2(x)$).
We can assume $Z=0$ or the solutions may not occur at the assumed $\pi$ intervals.