Question

How many solutions does tan^-1 3 have in interval [0, 2 π)

Answer 1

$\bf tan^{-1}(3)\iff tan^{-1}\left( \frac{\pm 3}{\pm 1} \right)=\theta \\\\\\ thus\qquad tan(\theta)=\cfrac{\pm 3}{\pm 1}\cfrac{\leftarrow opposite=y}{\leftarrow adjacent=x}$

now, the angle of θ, can only have a "y" value that is positive on, well, y is positive at 1st and 2nd quadrants

and "x" is positive only in 1st and 4th quadrants

now, that angle θ, can only have those two fellows, "y" and "x" to be positive, only in the 1st quadrant, and also both to be negative on the 3rd quadrant.

and that those two fellows, can also be both negative in the 3rd quadrant
3/1 = 3, and -3/-1 = 3

so, the solutions can only be "3", when both "y" and "x" are the same sign, and that only occurs on the 1st and 3rd quadrants