Find the general solution : `sec^2 2x=1-tan2x`

Explanation

To solve the equation \( \sec^2 2x = 1 - \tan 2x \), we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities.
We know that:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Thus, we can rewrite the left side of the equation:
\[ \sec^2 2x = 1 + \tan^2 2x \]
Substituting this into the equation gives:
\[ 1 + \tan^2 2x = 1 - \tan 2x \]

### Step 2: Simplify the equation.
Subtract 1 from both sides:
\[ \tan^2 2x = -\tan 2x \]
Rearranging gives:
\[ \tan^2 2x + \tan 2x = 0 \]

### Step 3: Factor the equation.
We can factor out \( \tan 2x \):
\[ \tan 2x (\tan 2x + 1) = 0 \]

### Step 4: Set each factor to zero.
This gives us two cases to solve:
1. \( \tan 2x = 0 \)
2. \( \tan 2x + 1 = 0 \) or \( \tan 2x = -1 \)

### Step 5: Solve the first case \( \tan 2x = 0 \).
The general solution for \( \tan \theta = 0 \) is:
\[ 2x = n\pi \quad \text{where } n \in \mathbb{Z} \]
Thus, dividing by 2:
\[ x = \frac{n\pi}{2} \]

### Step 6: Solve the second case \( \tan 2x = -1 \).
The general solution for \( \tan \theta = -1 \) is:
\[ 2x = n\pi - \frac{\pi}{4} \quad \text{where } n \in \mathbb{Z} \]
Dividing by 2 gives:
\[ x = \frac{n\pi}{2} - \frac{\pi}{8} \]

### Step 7: Combine the solutions.
The general solution for the equation \( \sec^2 2x = 1 - \tan 2x \) is:
\[ x = \frac{n\pi}{2} \quad \text{and} \quad x = \frac{n\pi}{2} - \frac{\pi}{8} \quad \text{where } n \in \mathbb{Z} \]