I have the function $$ \begin{align} f(x) = &\left(z × \cos(α) × (\cos(x + α - \tan(α)) + x × \sin(x + α - \tan(α))) - z × \cos\left(\frac{F}{z × \cos(α)}\right)\right)^2 \\ + &\left(z × \cos(α) × (\sin(x + α - \tan(α)) - x × \cos(x + α - \tan(α))) + z × \sin\left(\frac{F}{z × \cos(α)}\right)\right)^2 - F^2 \end{align} $$
where $F$, $Z$, and $α$ are positive real numbers. I want to find the simplest possible description of its roots, the places where $f(x) = 0$.
If I put the function into WolframScript as-is, it tells me that the equation cannot be solved. However, after plotting the function in GeoGebra, I noticed that one of the roots also happens to have a slope of zero.
Taking the derivative of $f(x)$ gives $$ f'(x) = 2 × z^2 × \cos(α) × \left(\cos(α) - \cos\left(α - \tan(α) + \frac{F}{z × \cos(α)} + x\right)\right) × x $$
WolframScript will find the roots to that expression, with the one I want working out to $x_1 = \tan(α) - \frac{F}{z × \cos(α)}$. Plugging that into the original function confirms that $f(x_1) = 0$.
I tried dividing $f(x)$ by $(x - x_1)^2$ to see if I could get it to simplify any, without success. Plotting the result does show that only two roots remain, but that's as far as I've gotten.
Is there a simplified function $g(x)$ with exactly two real roots which are equal to the two roots of $\frac{f(x)}{(x - x_1)^2}$?
I don't care about any other values of the hypothetical $g(x)$, just that it equals zero if and only if $\frac{f(x)}{(x - x_1)^2} = 0$ and that it is simpler to write out than $f(x)$.