### 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: I recognize that the equation is asking when the cosine of some angle in degrees is equal to 1, which happens only when the angle is a multiple of 360 degrees.
So I can rewrite the equation as
\[(k^2 + 6^2)^\circ = 360n\]
where $n$ is any integer.
Dividing both sides by 360, I get
\[\frac{k^2 + 6^2}{360} = n\]
I want to find the smallest positive values of $k$ that make this equation true, so I start by plugging in $n = 1$ and solving for $k$.
I get
\[k^2 + 6^2 = 360\]
\[k^2 = 360 - 36 = 324\]
\[k = \pm \sqrt{324} = \pm 18\]
Since $k$ has to be positive, I disregard the negative solution and keep $k = 18$ as one possible answer.
To find the next smallest value of $k$, I try $n = 2$ and repeat the same steps.
I get
\[k^2 + 6^2 = 720\]
\[k^2 = 720 - 36 = 684\]
\[k = \pm \sqrt{684} \approx \pm 26.15\]
Again, I only keep the positive solution, so $k \approx 26.15$ is another possible answer.
However, I notice that the problem asks for exact solutions, not approximate ones, so I wonder if there is a way to simplify $\sqrt{684}$.
I recall that $\sqrt{684} = \sqrt{4 \cdot 171} = 2\sqrt{171}$, and I check if 171 is a perfect square or has any perfect square factors.
I see that 171 is not a perfect square, but it is divisible by 9, which is a perfect square.
So I can write
\[\sqrt{171} = \sqrt{9 \cdot 19} = 3\sqrt{19}\]
Therefore, I can simplify $\sqrt{684}$ further as
\[\sqrt{684} = 2\sqrt{171} = 2 \cdot 3 \sqrt{19} = 6\sqrt{19}\]
This is the exact value of the second smallest positive solution for $k$.