Solve the equation for ` 0 le x le 2pi.`
`2 + 7 tan ^(2) theta = 3.25 sec ^(2) theta `

Answer

To solve the equation \( 2 + 7 \tan^2 \theta = 3.25 \sec^2 \theta \) for \( 0 \leq \theta \leq 2\pi \), we can follow these steps:

### Step 1: Rewrite the equation using the identity
We know that \( \sec^2 \theta = 1 + \tan^2 \theta \). Substituting this into the equation gives:
\[ 2 + 7 \tan^2 \theta = 3.25 (1 + \tan^2 \theta) \]

### Step 2: Expand the equation
Expanding the right side, we have:
\[ 2 + 7 \tan^2 \theta = 3.25 + 3.25 \tan^2 \theta \]

### Step 3: Rearrange the equation
Now, rearranging the equation to isolate terms involving \( \tan^2 \theta \):
\[ 7 \tan^2 \theta - 3.25 \tan^2 \theta = 3.25 - 2 \]
This simplifies to:
\[ (7 - 3.25) \tan^2 \theta = 1.25 \]
\[ 3.75 \tan^2 \theta = 1.25 \]

### Step 4: Solve for \( \tan^2 \theta \)
Dividing both sides by 3.75 gives:
\[ \tan^2 \theta = \frac{1.25}{3.75} = \frac{1}{3} \]

### Step 5: Take the square root
Taking the square root of both sides, we find:
\[ \tan \theta = \pm \frac{1}{\sqrt{3}} \]

### Step 6: Find the angles
The values of \( \theta \) for \( \tan \theta = \frac{1}{\sqrt{3}} \) are:
\[ \theta = \frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]
For \( n = 0 \):
\[ \theta = \frac{\pi}{6} \]
For \( n = 1 \):
\[ \theta = \frac{\pi}{6} + \pi = \frac{7\pi}{6} \]

The values of \( \theta \) for \( \tan \theta = -\frac{1}{\sqrt{3}} \) are:
\[ \theta = \frac{5\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]
For \( n = 0 \):
\[ \theta = \frac{5\pi}{6} \]
For \( n = 1 \):
\[ \theta = \frac{5\pi}{6} + \pi = \frac{11\pi}{6} \]

### Step 7: List the solutions
Thus, the solutions for \( 0 \leq \theta \leq 2\pi \) are:
\[ \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \]

### Final Answer
The final solutions are:
\[ \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \]