Step 1: Understanding the Question:
The given trigonometric equation involves \(\tan^2 \theta\) and \(\sec 2\theta\).
We need to find the general solution for \(\theta\).
Step 2: Key Formula or Approach:
Use the identity \(\sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta}\).
The general solution for \(\tan^2 \theta = \tan^2 \alpha\) is \(\theta = n\pi \pm \alpha\).
Step 3: Detailed Explanation:
Substitute the identity into the equation:
\[ \tan^2 \theta + \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta} = 1 \]
Let \(t^2 = \tan^2 \theta\). Then:
\[ t^2 + \frac{1 + t^2}{1 - t^2} = 1 \]
Multiply through by \((1 - t^2)\):
\[ t^2(1 - t^2) + 1 + t^2 = 1 - t^2 \]
\[ t^2 - t^4 + 1 + t^2 = 1 - t^2 \]
\[ 2t^2 - t^4 + 1 = 1 - t^2 \]
\[ t^4 - 3t^2 = 0 \]
\[ t^2(t^2 - 3) = 0 \]
Case 1: \(t^2 = 0 \Rightarrow \tan^2 \theta = 0 \Rightarrow \tan \theta = 0\).
The general solution is \(\theta = n\pi, n \in \mathbb{Z}\).
Case 2: \(t^2 - 3 = 0 \Rightarrow \tan^2 \theta = 3\).
Since \(3 = (\sqrt{3})^2 = \tan^2(\frac{\pi}{3})\), we have \(\tan^2 \theta = \tan^2(\frac{\pi}{3})\).
The general solution is \(\theta = n\pi \pm \frac{\pi}{3}, n \in \mathbb{Z}\).
Step 4: Final Answer:
Combining the solutions, we get \(\theta = n\pi\) or \(\theta = n\pi \pm \frac{\pi}{3}\).