The problem presented is:

$\cot^2 x = 1$

To solve for $x$, we can follow these steps:

1. Take the square root of both sides:

   $\cot x = \pm 1$

2. Now, consider the values of $x$ for which $\cot x = 1$ and $\cot x = -1$.

   • For $\cot x = 1$: The value of $x$ that satisfies $\cot x = 1$ within one period $[0, \pi]$ is $x = \frac{\pi}{4}$.
   • For $\cot x = -1$: The value of $x$ that satisfies $\cot x = -1$ within the same interval is $x = \frac{3\pi}{4}$.

3. Since cotangent is periodic with period $\pi$, the general solutions are:

   $x = \frac{\pi}{4} + n\pi \quad \text{and} \quad x = \frac{3\pi}{4} + n\pi$

   where $n$ is any integer.

Thus, the solutions to the equation $\cot^2 x = 1$ are:

$x = \frac{\pi}{4} + n\pi \quad \text{and} \quad x = \frac{3\pi}{4} + n\pi$

Would you like further details on this solution or explanations on related concepts?

Here are 5 related questions for deeper understanding:

1. What are the general properties of the cotangent function, especially its period and symmetry?
2. How does one derive the period of trigonometric functions like $\cot x$?
3. What is the significance of finding general solutions in trigonometric equations?
4. How would the solution change if the equation were $\cot^2 x = 2$ instead of $\cot^2 x = 1$?
5. How can we verify the solutions of trigonometric equations graphically?

Tip: When solving trigonometric equations, always consider the periodic nature of the functions to find all possible solutions within the desired interval.