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

Answer

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

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \sec^2 \theta = 1 + \tan^2 \theta \). Therefore, we can rewrite the left side of the equation:
\[ \sec^2 2x = 1 + \tan^2 2x \]
Substituting this into our 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 \( \tan 2x \):
\[ \tan 2x (\tan 2x + 1) = 0 \]
This gives us two cases to solve:
1. \( \tan 2x = 0 \)
2. \( \tan 2x + 1 = 0 \) or \( \tan 2x = -1 \)

### Step 4: Solve the first case \( \tan 2x = 0 \)
The general solution for \( \tan \theta = 0 \) is:
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]
Thus, for \( 2x \):
\[ 2x = n\pi \implies x = \frac{n\pi}{2} \]

### Step 5: Solve the second case \( \tan 2x = -1 \)
The general solution for \( \tan \theta = -1 \) is:
\[ \theta = -\frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
Thus, for \( 2x \):
\[ 2x = -\frac{\pi}{4} + n\pi \implies x = -\frac{\pi}{8} + \frac{n\pi}{2} \]

### Step 6: Combine the solutions
We have two general solutions:
1. \( x = \frac{n\pi}{2} \)
2. \( x = -\frac{\pi}{8} + \frac{n\pi}{2} \)

### Final Answer
The general solutions for the equation \( \sec^2 2x = 1 - \tan 2x \) are:
\[ x = \frac{n\pi}{2} \quad \text{and} \quad x = -\frac{\pi}{8} + \frac{n\pi}{2} \quad (n \in \mathbb{Z}) \]