Comparing populations using trigonometry. Given the function: 
$$
f(t) = 62 \cos \left( \frac{2π}{2⋅46}t \right)+138.
$$
The function above represents population #1.
If there was a population #2 that rises considerably when population #1 is under 100, what percentage of the time can one expect to see the higher population #2?
 A: I introduce this plot to visually describe my interpretation of the problem.

Here the big blue sinusoid represents population 1, and the small squiggly red line represents the places in time where we can "see" population 2, this being everywhere that population 1 is less than 100. So this reduces the question to the percentage of time that we can see population 2 versus the time that we cannot see population 2. Now by symmetry it is sufficient to consider this percentage for just the first half period on $0 \leq t \leq 46$. This ratio is marked out by the thick red and blue horizontal lines arbitrarily placed at the beginning of the plot (marked off as everything to the left of the vertical dotted line).
We proceed by finding the interval of time on $0 \leq t \leq 46$ where $f(t)<100$. First find the critical point $t$ where $f(t)=100$:
$$
\begin{align*}
f(t)=100  & \Rightarrow 62\cos\left( \frac{\pi}{46}t \right)+138=100\\
&\Rightarrow \cos\left( \frac{\pi}{46}t \right)=\frac{100-138}{62}=-\frac{19}{31} \\
&\Rightarrow \cos^{-1}\left( -\frac{19}{31} \right)=\frac{\pi}{46}t \\
&\Rightarrow t=\frac{46\cos^{-1}\left( -\frac{19}{31} \right)}{\pi} \approx 33.
\end{align*} 
$$
The conclusion here is that, over this first half period, population 1 is greater than or equal to $100$ on the time interval $0 \leq t \leq \frac{46\cos^{-1}\left( -\frac{19}{31} \right)}{\pi}$, and it falls below $100$ on the time interval $\frac{46\cos^{-1}\left( -\frac{19}{31} \right)}{\pi}<t\leq46$.
We take the ratio of times multiplied by 100 to get the percent of time that we can expect to see the higher population 2:
$$
\frac{46-\frac{46\cos^{-1}\left( -\frac{19}{31} \right)}{\pi}}{46} \cdot 100 \% \approx 29 \%
$$
We expect to see population 2 approximately $29 \%$ of the time.  
