Solve `sin^(2) theta gt cos^(2) theta`.

Explanation

To solve the inequality \( \sin^2 \theta > \cos^2 \theta \), we can follow these steps:

### Step 1: Rewrite the inequality
We start with the given inequality:
\[ \sin^2 \theta > \cos^2 \theta \]
This can be rewritten as:
\[ \sin^2 \theta - \cos^2 \theta > 0 \]

### Step 2: Use the Pythagorean identity
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). Therefore, we can express \( \sin^2 \theta \) in terms of \( \cos^2 \theta \):
\[ \sin^2 \theta = 1 - \cos^2 \theta \]
Substituting this into our inequality gives:
\[ (1 - \cos^2 \theta) - \cos^2 \theta > 0 \]
Simplifying this, we have:
\[ 1 - 2\cos^2 \theta > 0 \]

### Step 3: Rearranging the inequality
Rearranging the inequality, we get:
\[ 2\cos^2 \theta < 1 \]
Dividing both sides by 2:
\[ \cos^2 \theta < \frac{1}{2} \]

### Step 4: Taking the square root
Taking the square root of both sides (and remembering to consider both positive and negative roots), we find:
\[ |\cos \theta| < \frac{1}{\sqrt{2}} \]
This can be expressed as:
\[ -\frac{1}{\sqrt{2}} < \cos \theta < \frac{1}{\sqrt{2}} \]

### Step 5: Finding the angles
The cosine function is negative in the second quadrant and positive in the first quadrant. The angles corresponding to \( \cos \theta = \frac{1}{\sqrt{2}} \) are:
\[ \theta = \frac{\pi}{4} + 2n\pi \quad \text{and} \quad \theta = \frac{7\pi}{4} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, the inequality \( -\frac{1}{\sqrt{2}} < \cos \theta < \frac{1}{\sqrt{2}} \) corresponds to:
\[ \frac{\pi}{4} < \theta < \frac{3\pi}{4} \quad \text{and} \quad \frac{5\pi}{4} < \theta < \frac{7\pi}{4} \]

### Final Solution
The general solution for the inequality \( \sin^2 \theta > \cos^2 \theta \) is:
\[ \theta \in \left( \frac{\pi}{4} + 2n\pi, \frac{3\pi}{4} + 2n\pi \right) \cup \left( \frac{5\pi}{4} + 2n\pi, \frac{7\pi}{4} + 2n\pi \right) \quad \text{for } n \in \mathbb{Z} \]