General solution of `8tan^(2)((x)/(2))=1+secx` is

Options

A. `2npipmcos^(-1)((2)/(3)), ninZ`
B. `2npipmcos^(-1)((1)/(3)), ninZ`
C. `npipmcos^(-1)((2)/(3)), ninZ`
D. `npipmcos^(-1)((1)/(3)), ninZ`

Answer

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

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

### Step 2: Use the double angle identities
Using the double angle identities:
\[ \sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos x}{2} \quad \text{and} \quad \cos^2\left(\frac{x}{2}\right) = \frac{1 + \cos x}{2} \]
Substituting these into the equation gives:
\[ 8 \frac{\frac{1 - \cos x}{2}}{\frac{1 + \cos x}{2}} = 1 + \frac{1}{\cos x} \]
This simplifies to:
\[ 8 \frac{1 - \cos x}{1 + \cos x} = 1 + \frac{1}{\cos x} \]

### Step 3: Clear the fractions
To eliminate the fractions, we can multiply both sides by \( \cos x(1 + \cos x) \):
\[ 8 \cos x (1 - \cos x) = (1 + \cos x) \cos x \]
This simplifies to:
\[ 8 \cos x - 8 \cos^2 x = \cos x + \cos^2 x \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 8 \cos x - \cos x - 8 \cos^2 x - \cos^2 x = 0 \]
This simplifies to:
\[ 7 \cos x - 9 \cos^2 x = 0 \]

### Step 5: Factor the equation
Factoring out \( \cos x \):
\[ \cos x (7 - 9 \cos x) = 0 \]
This gives us two cases:
1. \( \cos x = 0 \)
2. \( 7 - 9 \cos x = 0 \)

### Step 6: Solve each case
**Case 1:** \( \cos x = 0 \)
This occurs at:
\[ x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \]

**Case 2:** \( 7 - 9 \cos x = 0 \)
Solving gives:
\[ \cos x = \frac{7}{9} \]
The general solution for this is:
\[ x = \cos^{-1}\left(\frac{7}{9}\right) + 2n\pi \quad \text{or} \quad x = -\cos^{-1}\left(\frac{7}{9}\right) + 2n\pi, \quad n \in \mathbb{Z} \]

### Final General Solution
Combining both cases, the general solution is:
\[ x = \frac{\pi}{2} + n\pi \quad \text{or} \quad x = \pm \cos^{-1}\left(\frac{7}{9}\right) + 2n\pi, \quad n \in \mathbb{Z} \]