Simplification ideas

Question

Looking for a neat simplification idea to be able to solve for $x$ analytically in the expression below: $$S=k\tan x-Bk^2\frac{1}{\cos^2x}$$

where $\{S,k,B\}\neq0$ and $\in \mathbb{R}^+.$

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Of course one could Taylor expand the ($\tan x$) and ($\frac{1}{\cos^2x}$) into ($\approx x+x^3/3$) and ($\approx 1+x^2$) respectively and reinsert into the main equation and solve the cubic polynomial, but aside from the fact that the obtained result in this way would only be an approximation valid around $0$, the analytic expressions would be too long and ugly. I figured there may be another more straightforward or reasonable way of attacking this.

Answer 1

Write $$\frac{1}{\cos^2(x)} = \sec^2(x)=\tan^2(x)+1.$$ The resulting equation is a quadratic in $\tan(x)$ which can be solved in terms of $S,k,B$. The $\arctan$ of this result gives $x$.

Answer 2

Hint:$$\dfrac{1}{\cos^2(x)}=\sec^2(x)=\tan^2(x)+1$$ Then use $\displaystyle u=\tan(x)$ in the resulting equation.