Let $A,B$ be constants, $\epsilon$ a small parameter, and $x\in(-\pi/2,\pi/2)$ s.t. $\epsilon\ll|\cos(x)|$. I would like to solve the equation $$ y = -A \tan x - \epsilon B \frac{1}{\cos(x)}\qquad\qquad (1) $$ for $x$ using asymptotic expansions. The unperturbed problem yields $$ x = \tan^{-1}(-\frac{y}{A}). $$ The Maclaurin series expansions are $$ \frac{1}{\cos x} = 1 + \frac{1}{2}x^2 + \frac{5}{24}x^4 + ... $$ and $$ \tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + ... $$ Equation (1) then becomes $$ y = -B\epsilon - Ax - \frac{1}{2}B\epsilon x^2 - \frac{1}{3}Ax^3 - \frac{5}{24}B\epsilon x^4 - \frac{2}{15}Ax^5 - ... $$ Using the power series ansatz $$ x_\epsilon = x_0 + \epsilon x_1 + \epsilon^2 x_3 + ... $$ equation (1) after some simplifications becomes $$ y = G_0 + \epsilon G_1 + \epsilon^2 G_2 + ... $$ with $$ G_0 = -Ax_0 - \frac{1}{3}Ax_0^3 - \frac{2}{15}Ax_0^5 - ...\\ G_1 = -Ax_1 - Ax_0^2x_1 - \frac{2}{3}Ax_0^4x_1 - B - \frac{1}{2}Bx_0^2 - \frac{5}{24}Bx_0^4 - ...\\ G_2 = -Ax_2 - Ax_2x_0^2 - \frac{2}{3}Ax_0^4x_2 - Ax_0x_1^2 - \frac{4}{3}Ax_0^3x_1^2 - Bx_0x_1 - \frac{5}{6}Bx_0^3x_1 - ... $$ This yields the $\mathcal{O}(\epsilon^0)$-Problem $$ y = -Ax_0 - \frac{1}{3}Ax_0^3 - \frac{2}{15}Ax_0^5 - ... $$ hence $$ -\frac{y}{A} = x_0 + \frac{1}{3}x_0^3 + \frac{2}{15}x_0^5 + ... = \tan(x_0) $$ which as expected gives $$ x_0 = \tan^{-1}(-\frac{y}{A}). $$ The $\mathcal{O}(\epsilon^1)$-Problem yields $$ 0 = Ax_1 + Ax_0^2x_1 + \frac{2}{3}Ax_0^4x_1 + B + \frac{1}{2}Bx_0^2 + \frac{5}{24}Bx_0^4 $$ i.e., $$ x_1 = -\frac{B + \frac{1}{2}Bx_0^2 + \frac{5}{24}Bx_0^4}{A + Ax_0^2 + \frac{2}{3}x_0^4}. $$ I will save the effort for copying the $x_2$ part of the expansion, because I am not too confident that I am on the right track.
EDIT: I have made a mistake when computing the $x_0$-Solution. Corrected that. Now, I am asking more about confirmation, I guess.
Since $x\neq \pm\frac \pi2$, multiply by $\cos(x)$ and you need to solve for $x$ the equation $$A \sin (x)+y \cos (x)+B \epsilon =0 \tag 1$$
Expanded around $x=-\tan ^{-1}\left(\frac{y}{A}\right)$, we have $$A \sin (x)+y \cos (x)=A \sqrt{1+\frac{y^2}{A^2}}\,\,\sum_{n=0}^\infty (-1)^n\frac {\left(x+\tan ^{-1}\left(\frac{y}{A}\right)\right)^{2n+1} }{(2n+1)! }$$
Let $$\alpha =A \sqrt{1+\frac{y^2}{A^2}} \qquad \text{and} \qquad \beta=\tan ^{-1}\left(\frac{y}{A}\right)$$ Using series reversion $$x=-\beta+t+\frac{t^3}{6}+\frac{3 t^5}{40}+\frac{5 t^7}{112}+O\left(t^{9}\right)\quad \text{where} \quad t=\frac{B \epsilon }{\alpha }$$