trig function restriction set notation The question is: 

Find the following expression that describes the location of each of the following values for $y=\tan(\theta)$, where $n \in I$ and $\theta $  is in radians.
  a) $\theta$-intercepts
  b) vertical-asymptotes

What does $n \in I$ mean, it makes sense for $I$ to refer to integers, but I know $\mathbb{Z}$ refers to integers, so what's going on?
My answers were 
a) $y= n\pi$ 
b) $y= n\pi \pm \frac{\pi}{2}$
Textbook answers showed the same but for...
b) $y=n\pi + \frac{\pi}{2}$
Why did they only show $+$, and not $\pm$, does it have something to do with this mysterious $x \in I$ restriction which doesn't seem to be anywhere on the internet.
 A: Evidently, the author is using $I$ for the set of integers rather than $\mathbb{Z}$.
Let's consider the graph of $y = \tan\theta$.

A particular solution of the equation $\tan\theta = 0$ is $\theta = 0$.  Since the tangent function $y = \tan\theta$ is periodic with period $\pi$ and the tangent function assumes each real value exactly once in each period, the set of all $\theta$-intercepts is $\theta = 0 + n\pi = n\pi, n \in \mathbb{Z}$, as you found.
Since 
$$y = \tan\theta = \frac{\sin\theta}{\cos\theta}$$
and there does exist a $\theta$ such that $\sin\theta = \cos\theta = 0$, the
tangent function has a vertical asymptote whenever $\cos\theta = 0$.  A particular solution of the equation $\cos\theta = 0$ is $\theta = \frac{\pi}{2}$.   Since any angle whose terminal side lies on the $y$-axis has $\cos\theta = 0$, the set of all solutions of the equation $\cos\theta = 0$ is $\theta = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$.  
Notice that if $m \in \mathbb{Z}$,
$$m\pi - \frac{\pi}{2} = m\pi - \pi + \frac{\pi}{2} = (m - 1)\pi + \frac{\pi}{2}$$
Thus, the set of $\theta$ such that 
$$\theta = m\pi - \frac{\pi}{2}, m \in \mathbb{Z}$$
is equal to the set of $\theta$ such that 
$$\theta = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$$
Just take $n = m - 1$.
Therefore, by writing $\theta = n\pi \pm \frac{\pi}{2}$, you are listing every solution twice.
A: When you write $n\pm\frac12$, every value is duplicated. Prefer $n+\frac12$.
