Kepler's equation: $M=E-\varepsilon\cdot \sin( E)$, you need to solve for $E$.
The inverse problem may be solved (according to wikipedia) as follows:
$E = \begin{cases} \displaystyle \sum_{n=1}^{\infty} {\frac{M^{\frac{n}{3}}}{n!}} \lim_{\theta \to 0} \left( \frac{\mathrm{d}^{\,n-1}}{\mathrm{d}\theta^{\,n-1}} \left( \frac{\theta}{ \sqrt[3]{\theta - \sin(\theta)} } ^n \right) \right) , & \epsilon = 1 \\ \displaystyle \sum_{n=1}^{\infty} { \frac{ M^n }{ n! } } \lim_{\theta \to 0} \left( \frac{\mathrm{d}^{\,n-1}}{\mathrm{d}\theta^{\,n-1}} \left( \frac{ \theta }{ \theta - \epsilon \cdot \sin(\theta)} ^n \right) \right) , & \epsilon \ne 1 \end{cases}$
which evaluates to:
$E = \begin{cases} \displaystyle M + \frac{1}{60} M^3 + \frac{1}{1400}M^5 + \cdots \ | \ x = ( 6 M )^\frac{1}{3} , & \epsilon = 1 \\ \\ \displaystyle \frac{1}{1-\epsilon} M - \frac{\epsilon}{( 1-\epsilon)^4 } \frac{M^3}{3!} + \frac{(9 \epsilon^2 + \epsilon)}{(1-\epsilon)^7 } \frac{M^5}{5!} + \cdots , & \epsilon \ne 1 \end{cases} $
The above is copied straight out of wikipedia. It's kind of ugly. Plus, the first equation is written in "x" instead of "M" and you can't be sure it's correct.
So instead, what I'd try is to take the first terms from the above, i.e. put $E_0 = M$ or $M/(1-\epsilon)$, and use Newton's method to iterate. That is, iterate with:
$E_{n+1} = E_n - \frac{M - E +\epsilon \sin E}{-1+\epsilon\cos(E)}$.
To test the method, run a bunch of random data through it, and make sure you test the fence posts (boundary conditions).
Obviously it is not as obvious as I thought!
The radial velocity curve is defined through 6 free parameters
$$V_r(t) = K\left(\cos(\omega + \nu) +e \cos \omega \right) + \gamma,$$
where $K$ is the semi-amplitude, $\gamma$ is the centre of mass radial velocity, $\omega$ is the usual angle defining the argument of the pericentre measured from the ascending node and $\nu$ is the true anomlay, which is a function of time, the fiducial time of pericentre passage $\tau$, the orbital period $p$ and the eccentricity $e$.
To proceed you estimate what all these parameters are - i.e. an initial guess.
Then, for each time $t_i$ of a data point in your RV curve you:
Calculate the mean anomaly
$$M(t) = \frac{2\pi}{p}(t - \tau),$$
Solve "Kepler's equation"
$$M(t) = E(t) - e \sin E(t)$$
numerically to give $E(t_i)$, the eccentric anomaly.
Use
$$\tan \frac{E(t)}{2} = \left(\frac{1+e}{1-e}\right)^{-1/2} \tan \frac{\nu(t)}{2}$$
to calculate the true anomaly $\nu(t_i)$.
Calculate $V_r(t_i)$
You then calculate some figure of merit (e.g. chi-squared) and go through an iterative process to adjust the parameters and optimise the fit of model to data.
A more sophisticated discussion can be found in this paper by Beauge et al.
Best Answer
I had a related question in which I basically derived that equation without me knowing.
For this I did assumed that the following relation between the radius, $r$, and the true anomaly, $\theta$, is correct, $$ r=\frac{a(1-e^2)}{1+e\cos{\theta}} $$ with $a$ the semi-major axis and $e$ the eccentricity.
By using Kepler's second and third law you can get an expression for time (since periapsis passage) as a function of the true anomaly, which looks like this: $$ t(\theta)=\sqrt{\frac{a^3}{\mu}}\left(2\tan^{-1}\left(\sqrt{\frac{1-e}{1+e}}\tan{\frac{\theta}{2}}\right)-\frac{e\sqrt{1-e^2}\sin{\theta}}{1+e\cos{\theta}}\right) $$ This equations is equivalent to Kepler's equation, because the mean and eccentric anomaly are defined as followed: $$ M=t\sqrt{\frac{\mu}{a^3}} $$ $$ \tan{\frac{\theta}{2}}=\sqrt{\frac{1+e}{1-e}}\tan{\frac{E}{2}} $$ A few substitutions are easy to see, namely how to get $M$ on the left hand side, by dividing by $\sqrt{\frac{a^3}{\mu}}$, and how to get the linear term of $E$, which allows use to get the following equation: $$ M=E-\frac{e\sqrt{1-e^2}\sin{\theta}}{1+e\cos{\theta}} $$ Proofing that the remaining term can be expressed as a product of the sine of $E$ and the eccentricity $e$ is harder, so $$ \sin{E}=\frac{\sqrt{1-e^2}\sin{\theta}}{1+e\ \cos{\theta}} $$
With the use of a temporary variable it is possible to proof this. This variable, lets call it $\alpha$, is defined as follows,
$$ \alpha = \tan \frac{\theta}{2} = \sqrt{\frac{1+e}{1-e}}\tan{\frac{E}{2}}. $$
This way the trigonometry terms in the fraction become:
$$ \sin \theta = \sin \left( 2 \tan^{-1} \alpha\right) = \frac{2\alpha}{1+\alpha^2} $$
$$ \cos \theta = \cos \left( 2 \tan^{-1} \alpha\right) = \frac{1-\alpha^2}{1+\alpha^2} $$
Substituting these back into the fraction yields:
$$ \frac{\sqrt{1-e^2}\sin{\theta}}{1+e\ \cos{\theta}} = \frac{\sqrt{1-e^2}\frac{2\alpha}{1+\alpha^2}}{1+e\frac{1-\alpha^2}{1+\alpha^2}} = \frac{2\sqrt{1-e^2}\alpha}{1+e+(1-e)\alpha^2} $$
Now by substituting in the expression for $\alpha$ which is an expression of $E$ gives:
$$ \frac{2\sqrt{1-e^2}\alpha}{1+e+(1-e)\alpha^2} = \frac{2\sqrt{(1+e)(1-e)}\sqrt{\frac{1+e}{1-e}}\tan{\frac{E}{2}}}{1+e+(1-e)\left(\sqrt{\frac{1+e}{1-e}}\tan{\frac{E}{2}}\right)^2} = \frac{2(1+e)\tan{\frac{E}{2}}}{(1+e)\left(1+\tan^2{\frac{E}{2}}\right)} $$
In this last expression the dependency on the eccentrically, $e$, can be removed, which yields,
$$ \frac{2\tan{\frac{E}{2}}}{1+\tan^2{\frac{E}{2}}} = \frac{2\frac{\sin(E/2)}{\cos(E/2)}}{1+\frac{\sin^2(E/2)}{\cos^2(E/2)}} = \frac{2\frac{\sin(E/2)}{\cos(E/2)}}{\frac{\cos^2(E/2)+\sin^2(E/2)}{\cos^2(E/2)}} = 2\sin\frac{E}{2}\cos\frac{E}{2}=\sin E $$
Therefore it can be shown that:
$$ \frac{\sqrt{1-e^2}\sin{\theta}}{1+e\ \cos{\theta}}=\sin{E} $$