### Instruction: There are infinitely many positive integers $k$ which satisfy the equation
\[\cos^2 (k^2 + 6^2)^\circ = 1.\]Enter the two smallest solutions, separated by commas.### Response: This problem involves a trigonometric equation and some algebra.
I notice that the cosine function is squared, which means that it will be positive regardless of the angle.
The only way that $\cos^2 \theta = 1$ is if $\cos \theta = \pm 1$.
This happens when $\theta$ is a multiple of $180^\circ$.
So I need to find values of $k$ such that $(k^2 + 6^2)^\circ$ is a multiple of $180^\circ$.
This means that $(k^2 + 6^2)$ must be a multiple of $180$.
I can rewrite this as $k^2 + 36 = 180n$, where $n$ is a positive integer.
Subtracting $36$ from both sides, I get $k^2 = 180n - 36$.
Taking the square root of both sides, I get $k = \pm \sqrt{180n - 36}$.
Since $k$ has to be a positive integer, I can ignore the negative sign and focus on finding values of $n$ that make the square root an integer.
I can try some small values of $n$ and see if they work.
If $n = 1$, then $k = \sqrt{180 - 36} = \sqrt{144} = 12$.
This is a valid solution, and it is the smallest one possible, since any smaller value of $n$ would make the square root negative.
If $n = 2$, then $k = \sqrt{360 - 36} = \sqrt{324} = 18$.
This is also a valid solution, and it is the second smallest one possible.
Any larger value of $n$ would make $k$ larger as well.
Therefore, the two smallest solutions for $k$ are $12$ and $18$.