Finding the Inverse A function $g$ is defined by $g(t) = 0.3(1 - \cos(2\pi t))$. 
For $n \in \mathbb{N}$, $I_n = [\frac{n}{2}, \frac{n+1}{2}]$. I have found that on this $n$-interval, $g$ is one-to-one meaning we can take the inverse of $g$ when restricted to this domain. 
For $n \in \mathbb{N}$, $h_n$ is the inverse of $g$ when restricted by the interval $I_n$. How can I find the formula for $h_n$?
I know that the formula depends on whether $n$ is even or odd, but I don't know how to find it.

The problem here is that when you try to invert the function $g(t)$, it will automatically appear to be in the interval $\left [ 0, \frac{1}{2}\right ]$ for even $n$, or $\left [ -\frac{1}{2}, 0\right ]$ for odd $n$.  To amend this, simply define a translation function $f_{even}: \left [ \frac{n}{2}, \frac{n+1}{2}\right ] \to \left[0, \frac{1}{2}\right ]$ or $f_{odd}: \left [ \frac{n}{2}, \frac{n+1}{2}\right ]\to \left [ -\frac{1}{2}, 0\right ] $ where we have
$$
f_{even}(x) \;\; =\;\; x - \frac{n}{2} \hspace{3pc} f_{odd}(x) \;\; =\;\; x - \frac{n+1}{2}.
$$
To get the correct bounds, re-interpret your function $g(t)$ to really be $(g\circ f_*)(t)$.  Then the inverse of this will be $\left (f_*^{-1}\circ g^{-1}\right )(t)$ where $* \in\{even, \; odd\}$.  
