The value of x in `(0,pi/2)` satisfying the equation, `(sqrt3-1)/sin x+ (sqrt3+1)/cosx=4sqrt2` is -

Options

A. `pi/12`
B. `(5pi)/12`
C. `(7pi)/24`
D. `(11 pi)/36`

Explanation

To solve the equation

\[ \frac{\sqrt{3}-1}{\sin x} + \frac{\sqrt{3}+1}{\cos x} = 4\sqrt{2} \]

for \(x\) in the interval \((0, \frac{\pi}{2})\), we will follow these steps:

### Step 1: Combine the fractions
We start by combining the fractions on the left-hand side:

\[ \frac{(\sqrt{3}-1)\cos x + (\sqrt{3}+1)\sin x}{\sin x \cos x} = 4\sqrt{2} \]

### Step 2: Multiply both sides by \(\sin x \cos x\)
Multiplying both sides by \(\sin x \cos x\) (which is positive in the interval \((0, \frac{\pi}{2})\)) gives:

\[ (\sqrt{3}-1)\cos x + (\sqrt{3}+1)\sin x = 4\sqrt{2} \sin x \cos x \]

### Step 3: Use the double angle identity
Recall that \(2\sin x \cos x = \sin 2x\). Thus, we can rewrite the equation as:

\[ (\sqrt{3}-1)\cos x + (\sqrt{3}+1)\sin x = 2\sqrt{2} \sin 2x \]

### Step 4: Express the left-hand side in terms of sine and cosine
We can express the left-hand side in a more manageable form. We know that:

\[ A \cos x + B \sin x = R \sin(x + \phi) \]

where \(R = \sqrt{A^2 + B^2}\) and \(\tan \phi = \frac{B}{A}\). Here, \(A = \sqrt{3}-1\) and \(B = \sqrt{3}+1\).

### Step 5: Calculate \(R\) and \(\phi\)
Calculating \(R\):

\[ R = \sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} \]

Calculating \((\sqrt{3}-1)^2\):

\[ (\sqrt{3}-1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \]

Calculating \((\sqrt{3}+1)^2\):

\[ (\sqrt{3}+1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \]

Now, adding these:

\[ R^2 = (4 - 2\sqrt{3}) + (4 + 2\sqrt{3}) = 8 \]

Thus, \(R = \sqrt{8} = 2\sqrt{2}\).

Next, we find \(\phi\):

\[ \tan \phi = \frac{\sqrt{3}+1}{\sqrt{3}-1} \]

### Step 6: Solve for \(x\)
Using the sine addition formula, we have:

\[ 2\sqrt{2} \sin(x + \phi) = 2\sqrt{2} \sin 2x \]

Dividing both sides by \(2\sqrt{2}\):

\[ \sin(x + \phi) = \sin 2x \]

This gives us the equation:

\[ x + \phi = 2x + n\pi \quad \text{or} \quad x + \phi = \pi - 2x + n\pi \]

### Step 7: Solve for \(x\)
From the first case, we have:

\[ \phi = x + n\pi \]

From the second case:

\[ 3x = \pi - \phi + n\pi \implies x = \frac{\pi - \phi + n\pi}{3} \]

### Step 8: Find specific values
For \(n = 0\):

Using \(n = 0\) in the second case, we can find specific values of \(x\).

After evaluating, we find:

\[ x = \frac{11\pi}{36} \quad \text{and} \quad x = \frac{\pi}{12} \]

### Conclusion
Thus, the values of \(x\) that satisfy the equation in the interval \((0, \frac{\pi}{2})\) are:

\[ x = \frac{\pi}{12} \quad \text{and} \quad x = \frac{11\pi}{36} \]