Question

Find the general solution of \( \tan^2 \theta = 1 \).

Hint:

For \( \tan^2 \theta = 1 \), consider both the positive and negative roots of the equation and solve for the general solution using periodicity of the tangent function.

Answer 1

Step 1: Solve the equation.
We are given the equation \( \tan^2 \theta = 1 \). Taking the square root of both sides, we get: \[ \tan \theta = \pm 1 \]

Step 2: Solve for \( \theta \).
The general solution for \( \tan \theta = 1 \) is: \[ \theta = \frac{\pi}{4} + n\pi \text{where} n \in \mathbb{Z} \] The general solution for \( \tan \theta = -1 \) is: \[ \theta = \frac{3\pi}{4} + n\pi \text{where} n \in \mathbb{Z} \]

Step 3: Combine the solutions.
Thus, the general solution is: \[ \theta = \frac{\pi}{4} + n\pi \text{or} \theta = \frac{3\pi}{4} + n\pi \text{where} n \in \mathbb{Z} \]

Final Answer: \[ \boxed{\theta = \frac{\pi}{4} + n\pi \text{ or } \theta = \frac{3\pi}{4} + n\pi \text{where} n \in \mathbb{Z}} \]