If I end up with $sin/cos/tan(n) = k$ , I can then solve for a value using each trigonometric function's inverse function, but that is only one of the values, that I can find... I looked up this subject, but I want confirmation on my notes:
let $ v \in \Bbb R: $
$sin (n) = k, n = sin^{-1}(k) \ and \ \ (\pi - sin^{-1}(k)) \pm 2\pi v $
$cos(n) = k, n = n = cos^{-1}(k) \ and \ \dots \pm 2\pi v$
$tan(n) = k, n = tan^{-1}(k) \pm \pi $
Correct me if I'm wrong, and do suggest a term to fill in the blanks, I'm going into trigonometry and I have doubts... Thank you very much!

According to the interval of definition and the range of trigonometric inverse functions we have that
$\sin n = k, n = \sin^{-1} k \lor n = \pi-\sin^{-1} k $
$\cos n= k, n = \cos^{-1} k \lor 2\pi -\cos^{-1} k$
$\tan n = k, n = \tan^{-1} k \lor \pi +\tan^{-1} k $
Note all values are up to a multiple of $2\pi$.
Try to derive each one thinking to the definition and selecting a value for n in each quadrant.
For example $\sin^{-1}x$ is defined for $x\in[-1,1]$ and has range $[-\pi/2,\pi/2]$. Thus - if n is in the range $\implies n = \sin^{-1} k$ - if n is not in the range $\implies n = \pi- \sin^{-1} k$