Rearranging equation $t = \frac{T}{2\pi} (\psi - \epsilon \sin \psi)$ in terms of $\psi$ I was playing around with the maths for orbits and trying to make a parametric equation that, well.. worked. I found a worksheet with parametrics with another variable ($\psi$), but I wanted to be able to put $t$ directly in somehow.
Is it possible to rearrange this equation
$$t = \frac{T}{2\pi} (\psi - \epsilon \sin \psi)$$
and make $\psi$ the subject?
$$\psi = ?$$
 A: In other words, isolating $\psi$ from the relation $$t = \frac{T}{2\pi} (\psi - \epsilon \sin \psi)$$ (which, as already said Thursday, cannot be explicitely inverted) means solving for $\psi$ $$f(\psi)=\psi - \epsilon \sin \psi -\frac{2\pi t}{T}=0$$ This implies numerical methods such as Newton which requires a "reasonable" starting guess (say $\psi_0$) of the solution.
Assuming that the solution searched for is such that $0\leq\psi\leq\pi$, we can approximate the sine function using $$\sin\psi \simeq a\psi(\pi-\psi)$$ So, to get an estimate,we solve first $$g(\psi)=a\epsilon \psi^2+  \left(1-a \pi \epsilon \right)\psi-\frac{2 \pi  t}{T}=0$$ which is just a quadratic with a positive solution corresponding to $$\psi=\frac{\sqrt{16 \pi  a t T+(\pi  a -2 )^2 T^2}+(\pi  a -2) T}{2 a T}$$ This will be $\psi_0$ from which we could start Newton iterations $$\psi_{n+1}=\psi_n-\frac{f(\psi_n)}{f'(\psi_n)}$$ 
With regard to the value of $a$, you could select $a=\frac{4}{\pi^2}$ if you want to reproduce $\sin(\frac{\pi}{2})=1$, $a=\frac{1}{\pi}$ if you want to reproduce the slopes at the bounds or $a=\frac{120}{\pi ^5}$ if you want to minimize $$\int_0^{\pi}\Big(\sin(x)-a x(\pi-x)\Big)^2 dx$$
Let us try using $a=\frac{4}{\pi^2}$, $T=100$,$\epsilon= \frac{1}{2}$,$t=40$ which lead to $\psi_0=2.73746$. Newton iterates will then be $2.71857$, $2.71854$ which is the solution for six significant figures.
A: Leibovici's answer is a more accurate way to do it. But if you just want some quick answers and are willing to sacrifice some accuracy, you can do a second order expansion for $\sin(\psi)$ and solve the resulting quadratic.
