The general solution of `8 tan^(2)""(x)/(2)=1+ sec x ` is

Options

A. ` x = 2n pi pm cos^(-1)""((-1)/(3))`
B. `x= 2 n pi pm (pi)/(6)`
C. `x = 2 n pi pm cos^(-1)""((1)/(3)) `
D. None of these

Answer

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

### Step 1: Simplify the equation
Start with the given equation:
\[ \frac{8 \tan^2 \left( \frac{x}{2} \right)}{2} = 1 + \sec x \]
This simplifies to:
\[ 4 \tan^2 \left( \frac{x}{2} \right) = 1 + \sec x \]

### Step 2: Rewrite \(\tan^2\) and \(\sec\) in terms of sine and cosine
We know that:
\[ \tan \left( \frac{x}{2} \right) = \frac{\sin \left( \frac{x}{2} \right)}{\cos \left( \frac{x}{2} \right)} \]
Thus,
\[ \tan^2 \left( \frac{x}{2} \right) = \frac{\sin^2 \left( \frac{x}{2} \right)}{\cos^2 \left( \frac{x}{2} \right)} \]
Also, recall that:
\[ \sec x = \frac{1}{\cos x} \]
Substituting these into the equation gives:
\[ 4 \cdot \frac{\sin^2 \left( \frac{x}{2} \right)}{\cos^2 \left( \frac{x}{2} \right)} = 1 + \frac{1}{\cos x} \]

### Step 3: Use half-angle identities
Using the half-angle identities:
\[ \sin^2 \left( \frac{x}{2} \right) = \frac{1 - \cos x}{2} \]
\[ \cos^2 \left( \frac{x}{2} \right) = \frac{1 + \cos x}{2} \]
Substituting these into the equation:
\[ 4 \cdot \frac{\frac{1 - \cos x}{2}}{\frac{1 + \cos x}{2}} = 1 + \frac{1}{\cos x} \]
This simplifies to:
\[ 4 \cdot \frac{1 - \cos x}{1 + \cos x} = 1 + \frac{1}{\cos x} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 4(1 - \cos x) \cdot \cos x = (1 + \cos x)(1 + \cos x) \]
Expanding both sides:
\[ 4\cos x - 4\cos^2 x = 1 + 2\cos x + \cos^2 x \]

### Step 5: Rearranging the equation
Rearranging terms leads to:
\[ 4\cos^2 x - 6\cos x + 1 = 0 \]

### Step 6: Solve the quadratic equation
Let \( y = \cos x \). The equation becomes:
\[ 4y^2 - 6y + 1 = 0 \]
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \]
Calculating the discriminant:
\[ = \frac{6 \pm \sqrt{36 - 16}}{8} = \frac{6 \pm \sqrt{20}}{8} = \frac{6 \pm 2\sqrt{5}}{8} = \frac{3 \pm \sqrt{5}}{4} \]

### Step 7: Find the general solution
Thus, we have:
\[ \cos x = \frac{3 + \sqrt{5}}{4} \quad \text{or} \quad \cos x = \frac{3 - \sqrt{5}}{4} \]
Let \( \alpha_1 = \cos^{-1} \left( \frac{3 + \sqrt{5}}{4} \right) \) and \( \alpha_2 = \cos^{-1} \left( \frac{3 - \sqrt{5}}{4} \right) \).

The general solutions are:
\[ x = 2n\pi \pm \alpha_1 \quad \text{and} \quad x = 2n\pi \pm \alpha_2 \quad (n \in \mathbb{Z}) \]

### Final Answer
The general solution of the equation is:
\[ x = 2n\pi \pm \cos^{-1} \left( \frac{3+\sqrt{5}}{4} \right) \quad \text{and} \quad x = 2n\pi \pm \cos^{-1} \left( \frac{3-\sqrt{5}}{4} \right) \quad (n \in \mathbb{Z}) \]