If `cos (pi/12) = (sqrt(2) + sqrt(6))/(4)`, then all `x in (0,pi/2)` such that `(sqrt(3)-1)/(sin x) + (sqrt(3)+1)/(cos x) = 4sqrt(2)`, then find x.

Answer

To solve the problem step by step, we need to find all values of \( x \) in the interval \( (0, \frac{\pi}{2}) \) such that:

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

### Step 1: Rewrite the equation

We start with the equation:

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

To combine the fractions, we take the common denominator \( \sin x \cos x \):

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

### Step 2: Cross-multiply

Cross-multiplying gives us:

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

### Step 3: Rearranging the equation

Rearranging the equation, we have:

\[ (\sqrt{3}-1)\cos x + (\sqrt{3}+1)\sin x - 4\sqrt{2} \sin x \cos x = 0 \]

### Step 4: Substitute values

We know from the problem that \( \cos \frac{\pi}{12} = \frac{\sqrt{2} + \sqrt{6}}{4} \). We can use this information later if needed, but for now, we will continue with the equation.

### Step 5: Use the identity

We can use the identity \( 2 \sin x \cos x = \sin(2x) \):

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

### Step 6: Factor and simplify

We can express the left-hand side in terms of sine and cosine:

Let \( A = \sqrt{3}-1 \) and \( B = \sqrt{3}+1 \):

\[ A \cos x + B \sin x = 2\sqrt{2} \sin(2x) \]

### Step 7: Find the angle

Using the sine addition formula, we can express \( A \cos x + B \sin x \) as:

\[ R \sin(x + \phi) \]

Where \( R = \sqrt{A^2 + B^2} \) and \( \tan \phi = \frac{B}{A} \).

Calculating \( R \):

\[ A^2 = (\sqrt{3}-1)^2 = 4 - 2\sqrt{3} \]
\[ B^2 = (\sqrt{3}+1)^2 = 4 + 2\sqrt{3} \]
\[ R^2 = A^2 + B^2 = (4 - 2\sqrt{3}) + (4 + 2\sqrt{3}) = 8 \]
\[ R = 2\sqrt{2} \]

### Step 8: Set up the equation

Now we have:

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

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

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

### Step 9: Solve for \( x \)

This gives us two cases:

1. \( x + \phi = 2x + 2k\pi \)
2. \( x + \phi = \pi - 2x + 2k\pi \)

For the first case:

\[ \phi = x + 2k\pi \implies x = \phi - 2k\pi \]

For the second case:

\[ x + \phi = \pi - 2x + 2k\pi \implies 3x = \pi - \phi + 2k\pi \implies x = \frac{\pi - \phi + 2k\pi}{3} \]

### Step 10: Find specific values

We will find \( \phi \) using \( \tan \phi = \frac{B}{A} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \).

Calculating \( \phi \) gives us:

\[ \phi = \frac{\pi}{12} \]

Thus, we can substitute back to find \( x \):

1. From the first case, \( x = \frac{\pi}{12} - 2k\pi \) (not in \( (0, \frac{\pi}{2}) \)).
2. From the second case, \( x = \frac{\pi - \frac{\pi}{12}}{3} = \frac{11\pi}{36} \).

### Final Answer

The value of \( x \) in the interval \( (0, \frac{\pi}{2}) \) is:

\[ \boxed{\frac{\pi}{12}} \]