Harvard-MIT November Tournament

Solution:
If $\sin \theta$, $\cos \theta$, and $\tan \theta$ are in a geometric progression, then the product of two must equal the square of the third. Using this criterion, we have 3 cases.

- Case 1: $\sin \theta \cdot \tan \theta = \cos^{2} \theta$. This implies that $\left(\sin^{2} \theta\right) = \left(\cos^{3} \theta\right)$. Writing $\sin^{2} \theta$ as $1-\cos^{2} \theta$ and letting $\cos \theta = x$, we have that $x^{3} + x^{2} - 1 = 0$. We wish to find the number of solutions of this where $|x| \leq 1$. Clearly $-1$ is not a root. If $-1 < x \leq 0$, we have that $x^{2} + x^{3} \leq x^{2} < 1$ so $x^{3} + x^{2} - 1 < 0$ and there are no roots. If $0 < x \leq 1$, then $x^{3} + x^{2} - 1$ is a strictly increasing function. Since it has value $-1$ at $x=0$ and value $1$ at $x=1$, there is exactly one root between $0$ and $1$, non-inclusive. There are 2 values of $\theta$ such that $\cos \theta$ equals this root, and thus, two solutions in this case.

- Case 2: $\sin \theta \cdot \cos \theta = \tan^{2} \theta$. This implies that $\cos^{3} \theta = \sin \theta$. To find the number of solutions in this case, we can analyze the graphs of the functions in different ranges. Note that from $\theta = 0$ to $\theta = \frac{\pi}{2}$, $\cos^{3} \theta$ decreases strictly from $1$ to $0$ while $\sin \theta$ increases strictly from $0$ to $1$. Hence, there is one solution in this range. By a similar argument, a solution exists between $\theta = \pi$ and $\theta = \frac{3\pi}{2}$. In the intervals $\left[\frac{\pi}{2}, \pi\right]$ and $\left[\frac{3\pi}{2}, 2\pi\right]$, we have that one function is negative and the other is positive, so there are no solutions. Thus, there are two solutions in this case.

- Case 3: $\cos \theta \cdot \tan \theta = \sin^{2} \theta$. This implies that $\sin \theta = \sin^{2} \theta$, so $\sin \theta = 0, 1$. Clearly the only solutions of these have $\theta$ as an integer multiple of $\frac{\pi}{2}$. Thus, there are no pertinent solutions in this case.

We can see that the solutions for the first two cases are mutually exclusive. Hence, there are $4$ solutions in total.