Well, I have the equality:
$$\theta+2\epsilon\sin\left(\theta\right)+\frac{3}{2}\cdot\epsilon^2\cdot\left(\theta+\cos\left(\theta\right)\sin\left(\theta\right)\right)\approx\alpha\cdot\text{n}\tag1$$
How can I solve this equation for $\theta$? For all the variables is know that they are real and positive.
Background of the problem:
I need to solve this for $\theta$:
$$\frac{\text{n}}{\text{A}}\int_0^{2\pi}\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=\int_0^\theta\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x\tag2$$
Now, for the LHS I got:
$$\frac{\text{n}}{\text{A}}\int_0^{2\pi}\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=\frac{1}{\text{A}}\cdot\frac{2\pi}{\sqrt{\left(1-\epsilon^2\right)^3}}\cdot\text{n}=\alpha\cdot\text{n}\tag3$$
Now, for the RHS:
$$\int_0^\theta\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=\sum_{\text{k}=0}^\infty\epsilon^\text{k}\left(1+\text{k}\right)\int_0^\theta\cos^\text{k}\left(x\right)\space\text{d}x\color{red}{\approx}$$ $$\sum_{\text{k}=0}^2\epsilon^\text{k}\left(1+\text{k}\right)\int_0^\theta\cos^\text{k}\left(x\right)\space\text{d}x=\theta+2\epsilon\sin\left(\theta\right)+\frac{3}{2}\cdot\epsilon^2\cdot\left(\theta+\cos\left(\theta\right)\sin\left(\theta\right)\right)\tag4$$
Where $0<\theta<2\pi$, so the error (in the red approximation sign) is:
$$0<\text{E}_3=\left|\sum_{\text{k}=3}^\infty\epsilon^\text{k}\left(1+\text{k}\right)\theta\right|<\frac{2\pi\epsilon^3\left(4-3\epsilon\right)}{\left(\epsilon-1\right)^2}\tag5$$
Let $$ \int_0^\theta\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=y(\theta) $$
The inverse of this function reads $$ \begin{aligned} \theta(y)&=y\\ &+2 \sin (y)\epsilon\\ &+\frac12\left(5 \sin (y) \cos (y)-3 y\right)\epsilon^2\\ &+\frac{1}{12} (-3 \sin (y)+13 \sin (3 y)-36 y \cos (y))\epsilon^3\\ &+\frac{1}{96} (36 y-44 \sin (2 y)+103 \sin (4 y)-360 y \cos (2 y))\epsilon^4+\cdots \end{aligned} $$
To derive the form of the coefficients, you need to use the identity $$ \begin{aligned} \int_0^{g(z)} f(x) \mathrm dx&=\int_0^{g(0)} f(x) \mathrm dx\\ &+z f(g(0))g'(0)\\ &+\frac12z^2\left(g'(0)^2f'(g(0))+f(g(0))g''(0)\right)\\ &+\frac16z^3\left(g'(0)^3f''(g(0))+3g'(0)g''(0)f'(g(0))+g'''(0)f(g(0))\right)+\cdots \end{aligned} $$ as given by Leibniz' formula. The general term is hopefully obvious from these first few terms.
In your particular case, $z\to\epsilon$, $g\to\theta$, and $$ y\equiv\frac{\text{n}}{\text{A}}\int_0^{2\pi}\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=\frac{\text{n}}{\text{A}}\frac{2 \pi }{\left(1-\epsilon ^2\right)^{3/2}} $$