The most general solution of the equation
`8"tan"^(2) (theta)/(2) = 1 +"sec" theta,` is

Options

A. `theta = 2n pi +-"cos"^(-1) ((1)/(3))`
B. `theta = 2 n pi +-(pi)/(6)`
C. `theta = 2 n pi +- "cos"^(-1)((-1)/(3))`
D. none of these

Answer

To solve the equation \( 8 \tan^2 \left( \frac{\theta}{2} \right) = 1 + \sec \theta \), we will follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \sec \theta = \frac{1}{\cos \theta} \]
Thus, we can rewrite the equation as:
\[ 8 \tan^2 \left( \frac{\theta}{2} \right) = 1 + \frac{1}{\cos \theta} \]

### Step 2: Use the identity for \(\cos \theta\)
Using the identity:
\[ \cos \theta = \frac{1 - \tan^2 \left( \frac{\theta}{2} \right)}{1 + \tan^2 \left( \frac{\theta}{2} \right)} \]
we can substitute this into our equation. The equation becomes:
\[ 8 \tan^2 \left( \frac{\theta}{2} \right) = 1 + \frac{1}{\frac{1 - \tan^2 \left( \frac{\theta}{2} \right)}{1 + \tan^2 \left( \frac{\theta}{2} \right)}} \]

### Step 3: Simplifying the right side
By simplifying the right side, we get:
\[ 1 + \frac{1 + \tan^2 \left( \frac{\theta}{2} \right)}{1 - \tan^2 \left( \frac{\theta}{2} \right)} = \frac{(1 - \tan^2 \left( \frac{\theta}{2} \right)) + (1 + \tan^2 \left( \frac{\theta}{2} \right))}{1 - \tan^2 \left( \frac{\theta}{2} \right)} = \frac{2}{1 - \tan^2 \left( \frac{\theta}{2} \right)} \]

### Step 4: Set the two sides equal
Now, we equate the two sides:
\[ 8 \tan^2 \left( \frac{\theta}{2} \right) = \frac{2}{1 - \tan^2 \left( \frac{\theta}{2} \right)} \]

### Step 5: Cross-multiply
Cross-multiplying gives:
\[ 8 \tan^2 \left( \frac{\theta}{2} \right) (1 - \tan^2 \left( \frac{\theta}{2} \right)) = 2 \]
Expanding this results in:
\[ 8 \tan^2 \left( \frac{\theta}{2} \right) - 8 \tan^4 \left( \frac{\theta}{2} \right) = 2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 8 \tan^4 \left( \frac{\theta}{2} \right) - 8 \tan^2 \left( \frac{\theta}{2} \right) + 2 = 0 \]

### Step 7: Let \( p = \tan^2 \left( \frac{\theta}{2} \right) \)
Letting \( p = \tan^2 \left( \frac{\theta}{2} \right) \), we rewrite the equation as:
\[ 8p^2 - 8p + 2 = 0 \]

### Step 8: Simplifying the quadratic equation
Dividing the entire equation by 2:
\[ 4p^2 - 4p + 1 = 0 \]

### Step 9: Factoring the quadratic
Factoring gives:
\[ (2p - 1)^2 = 0 \]
Thus, we find:
\[ 2p - 1 = 0 \implies p = \frac{1}{2} \]

### Step 10: Back-substituting for \( \tan^2 \left( \frac{\theta}{2} \right) \)
Substituting back, we have:
\[ \tan^2 \left( \frac{\theta}{2} \right) = \frac{1}{2} \]
Taking the square root:
\[ \tan \left( \frac{\theta}{2} \right) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]

### Step 11: Finding \( \theta \)
Thus, we have:
\[ \frac{\theta}{2} = \tan^{-1} \left( \frac{\sqrt{2}}{2} \right) + n\pi \quad \text{or} \quad \frac{\theta}{2} = \tan^{-1} \left( -\frac{\sqrt{2}}{2} \right) + n\pi \]
This gives:
\[ \theta = 2\tan^{-1} \left( \frac{\sqrt{2}}{2} \right) + 2n\pi \quad \text{or} \quad \theta = 2\tan^{-1} \left( -\frac{\sqrt{2}}{2} \right) + 2n\pi \]

### Step 12: Final solution
The general solution is:
\[ \theta = 2n\pi \pm \cos^{-1} \left( \frac{1}{3} \right) \]