Given an ellipse of parametric equations:
$$
(x,y) = (a\cos\alpha,b\sin\alpha)
\quad \quad \quad \text{with} \; \alpha \in [0,2\pi)
$$
the length of an arc between $\alpha=0$ and $\alpha=\beta$ is defined as:
$$
\mathcal{L}(\beta) := \int_0^{\beta} \underbrace{\sqrt{(a\sin\alpha)^2+(b\cos\alpha)^2}}_{f(\alpha)}\,\text{d}\alpha\,.
$$
Therefore, set $\Delta\beta \equiv 2\pi/n$, expanding $\mathcal{L}$ into first order Taylor series:
$$
\mathcal{L}(\beta+\Delta\beta) \approx \mathcal{L}(\beta) + \mathcal{L}'(\beta)\,\Delta\beta
$$
it's possible consider the following numerical procedure (Euler's method):
$$
\begin{cases}
\beta_0 = 0 \\
\mathcal{L}_0 = \dots \\
\end{cases}
\quad \quad \quad
\begin{cases}
\beta_i = \beta_{i-1} + \Delta\beta \\
\mathcal{L}_i = \mathcal{L}_{i-1} + f(\beta_{i-1})\,\Delta\beta \\
\end{cases}
\quad \quad \quad
i = 1,2,\dots,n
$$
through which we can achieve our goal in two steps:
let $\mathcal{L}_0 = 0$, it's possible to calculate the length of the ellipse $\mathcal{L}_e\equiv\mathcal{L}_n$;
let $\mathcal{L}_0 = -\lambda\,\mathcal{L}_e$ with $0 < \lambda < 1$, it's possible to calculate the desidered angle $\beta_{\lambda}\equiv\frac{\beta_{m-1}+\beta_m}{2}$, where $m<n$ is the index for which it's possible to stop the procedure when $\mathcal{L}_{m-1}\cdot\mathcal{L}_m<0$.
Finally, the angle $0 \le \gamma_{\lambda} < 2\pi$ between the vectors $(1,0)$ and $(x,y) = (a\cos\beta_{\lambda},b\sin\beta_{\lambda})$ is:
$$
\gamma_{\lambda} =
\begin{cases}
\arccos\left(\frac{x}{\sqrt{x^2+y^2}}\right) & \text{if} \; y \ge 0 \\
2\pi - \arccos\left(\frac{x}{\sqrt{x^2+y^2}}\right) & \text{if} \; y<0 \\
\end{cases}
$$
where only in the particular case of the circle $a=b$ we have $\gamma_{\lambda}=\beta_{\lambda}$ for all $\beta_{\lambda} \in [0,2\pi)$.
