Find the number of solutions to the equation
\[\tan (5 \pi \cos \theta) = \cot (5 \pi \sin \theta)\]where $\theta \in (0, 2 \pi).$
From the given equation,
\[\tan (5 \pi \cos \theta) = \frac{1}{\tan (5 \pi \sin \theta)},\]so $\tan (5 \pi \cos \theta) \tan (5 \pi \sin \theta) = 1.$

Then from the angle addition formula,
\begin{align*}
\cot (5 \pi \cos \theta + 5 \pi \sin \theta) &= \frac{1}{\tan (5 \pi \cos \theta + 5 \pi \sin \theta)} \\
&= \frac{1 - \tan (5 \pi \cos \theta) \tan (5 \pi \sin \theta)}{\tan (5 \pi \cos \theta) + \tan (5 \pi \sin \theta)} \\
&= 0.
\end{align*}Hence, $5 \pi \cos \theta + 5 \pi \sin \theta$ must be an odd multiple of $\frac{\pi}{2}.$  In other words,
\[5 \pi \cos \theta + 5 \pi \sin \theta = (2n + 1) \cdot \frac{\pi}{2}\]for some integer $n.$  Then
\[\cos \theta + \sin \theta = \frac{2n + 1}{10}.\]Using the angle addition formula, we can write
\begin{align*}
\cos \theta + \sin \theta &= \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \sin \theta \right) \\
&= \sqrt{2} \left( \sin \frac{\pi}{4} \cos \theta + \cos \frac{\pi}{4} \sin \theta \right) \\
&= \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right).
\end{align*}so
\[\sin \left( \theta + \frac{\pi}{4} \right) = \frac{2n + 1}{10 \sqrt{2}}.\]Thus, we need
\[\left| \frac{2n + 1}{10 \sqrt{2}} \right| \le 1.\]The integers $n$ that work are $-7,$ $-6,$ $-5,$ $\dots,$ $6,$ giving us a total of 14 possible values of $n.$  Furthermore, for each such value of $n,$ the equation
\[\sin \left( \theta + \frac{\pi}{4} \right) = \frac{2n + 1}{10 \sqrt{2}}.\]has exactly two solutions in $\theta.$  Therefore, there are a total of $\boxed{28}$ solutions $\theta.$
28
Level 5
Precalculus
Problem: Find the number of solutions to the equation
\[\tan (5 \pi \cos \theta) = \cot (5 \pi \sin \theta)\]where $\theta \in (0, 2 \pi).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation \(\tan (5 \pi \cos \theta) = \cot (5 \pi \sin \theta)\), we first rewrite the cotangent in terms of tangent: \(\cot x = \frac{1}{\tan x}\). This gives us the equation \(\tan (5 \pi \cos \theta) = \frac{1}{\tan (5 \pi \sin \theta)}\). 

Cross-multiplying leads to the equation \(\tan (5 \pi \cos \theta) \tan (5 \pi \sin \theta) = 1\). This implies that the angles \(5 \pi \cos \theta\) and \(5 \pi \sin \theta\) must satisfy the condition that their tangent values multiply to 1. 

The general solution for \(\tan A \tan B = 1\) occurs when \(A + B = \frac{\pi}{2} + n\pi\) for any integer \(n\). Thus, we can set up the equation:
\[
5 \pi \cos \theta + 5 \pi \sin \theta = \frac{\pi}{2} + n\pi.
\]
Dividing through by \(5\pi\) simplifies this to:
\[
\cos \theta + \sin \theta = \frac{1}{10} + \frac{n}{5}.
\]

Next, we know that \(\cos \theta + \sin \theta\) can be expressed as \(\sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right)\). The maximum value of \(\cos \theta + \sin \theta\) is \(\sqrt{2}\) and the minimum value is \(-\sqrt{2}\). Therefore, we need to find the integer values of \(n\) such that:
\[
-\sqrt{2} \leq \frac{1}{10} + \frac{n}{5} \leq \sqrt{2}.
\]

Calculating the bounds:
1. For the lower bound:
   \[
   -\sqrt{2} \leq \frac{1}{10} + \frac{n}{5} \implies n \geq 5(-\sqrt{2} - \frac{1}{10}) \approx -7.5 \implies n \geq -7.
   \]
2. For the upper bound:
   \[
   \frac{1}{10} + \frac{n}{5} \leq \sqrt{2} \implies n \leq 5(\sqrt{2} - \frac{1}{10}) \approx 6.5 \implies n \leq 6.
   \]

Thus, \(n\) can take integer values from \(-7\) to \(6\), which gives us \(14\) possible values for \(n\).

For each value of \(n\), the equation \(\cos \theta + \sin \theta = \frac{1}{10} + \frac{n}{5}\) can be solved for \(\theta\). Each solution corresponds to two angles in the interval \((0, 2\pi)\) due to the periodic nature of sine and cosine. Therefore, for each of the \(14\) values of \(n\), there are \(2\) solutions for \(\theta\).

Multiplying the number of \(n\) values by the number of solutions per \(n\):
\[
14 \times 2 = 28.
\]

Thus, the final answer is \(28\).<|END_OF_LATENT|> The final answer is $28$.