Confused About Equation Rearrangment Full disclosure; I have a very very basic understanding of higher math. I apologize if this is stupid / has been answered. I don't really know how to even format my question.
I have a formula used in ray casting off two mirrors. The formula is:
$$
z = \tan (d) (a+b \sec c)
$$
I need to rewrite this equation to solve for $d$.
I have used Wolfram Alpha for this, and it returns:
Formula
I don't understand what $n$ is, it wasn't a value in the input. I also don't have the knowledge / skills to rewrite this myself, or any idea of where to start.
Thanks!
 A: Is $d$ an angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$?  If so, you don't need the $n$.  
Divide both sides by $a+b\sec c$:
$$\frac{z}{a+b\sec c} = \tan d$$
And apply the inverse tangent function.  It satisfies $\tan^{-1}( \tan d) = d$ for all $-\frac{\pi}{2}$ and $\frac{\pi}{2}$: $$\tan^{-1}(\frac{z}{a+b\sec c}) = d$$ Given the values $z,a,b,c$ there are actually infinitely many numbers $d$ with the property that $\frac{z}{a+b\sec c} = \tan d$.  Indeed, let $$\tan^{-1}(\frac{z}{a+b\sec c})$$ be one such number.  Then $$\tan^{-1}(\frac{z}{a+b\sec c})+\pi$$$$ \tan^{-1}(\frac{z}{a+b\sec c})-\pi$$ $$\tan^{-1}(\frac{z}{a+b\sec c}) - 2\pi $$ $$\tan^{-1}(\frac{z}{a+b\sec c}) + 2\pi$$ and so on are all of the rest of the solutions to that equation.  Usually this is just written as $$\tan^{-1}(\frac{z}{a+b\sec c}) + \pi n$$ where $n$ is an integer, i.e. one of the numbers $\{... -2, -1, 0, 1, 2, ... \}$.
A: As $\tan x = \tan (x + \pi n)$, where $n\in\mathbb{Z}$, if you have $\tan d = \frac{z}{a + b\sec c}$ then all values $d$ satisfies this equation are gives by
$$
d = \tan^{-1}\left(\frac{z}{a + b\sec c}\right) + \pi n, \quad n\in \mathbb{Z}.
$$
