Solve the equation , ` sqrt(3) sin 2A = sin 2B and sqrt(3)sin^(2)A + sin^(2)B=(1)/(2)(sqrt(3)-1)`.

Answer

To solve the equations \( \sqrt{3} \sin 2A = \sin 2B \) and \( \sqrt{3} \sin^2 A + \sin^2 B = \frac{1}{2}(\sqrt{3}-1) \), we will follow these steps:

### Step 1: Rewrite the first equation
We start with the first equation:
\[ \sqrt{3} \sin 2A = \sin 2B \]
This can be rearranged to:
\[ \sqrt{3} \sin 2A - \sin 2B = 0 \]
Let’s denote this as Equation (1).

### Step 2: Rewrite the second equation
Next, we look at the second equation:
\[ \sqrt{3} \sin^2 A + \sin^2 B = \frac{1}{2}(\sqrt{3}-1) \]
We can denote this as Equation (2).

### Step 3: Use double angle identities
Using the double angle identity, we know:
\[ \sin^2 A = \frac{1 - \cos 2A}{2} \quad \text{and} \quad \sin^2 B = \frac{1 - \cos 2B}{2} \]
Substituting these into Equation (2):
\[ \sqrt{3} \left(\frac{1 - \cos 2A}{2}\right) + \left(\frac{1 - \cos 2B}{2}\right) = \frac{1}{2}(\sqrt{3}-1) \]
Multiplying through by 2 to eliminate the fractions:
\[ \sqrt{3}(1 - \cos 2A) + (1 - \cos 2B) = \sqrt{3} - 1 \]
This simplifies to:
\[ \sqrt{3} - \sqrt{3} \cos 2A + 1 - \cos 2B = \sqrt{3} - 1 \]
Rearranging gives:
\[ -\sqrt{3} \cos 2A - \cos 2B = -2 \]
Multiplying through by -1:
\[ \sqrt{3} \cos 2A + \cos 2B = 2 \]
Let’s denote this as Equation (3).

### Step 4: Square both equations
Now we have:
1. \( \sqrt{3} \sin 2A - \sin 2B = 0 \) (Equation 1)
2. \( \sqrt{3} \cos 2A + \cos 2B = 2 \) (Equation 3)

Next, we square both equations:
From Equation (1):
\[ 3 \sin^2 2A - 2\sqrt{3} \sin 2A \sin 2B + \sin^2 2B = 0 \]
From Equation (3):
\[ 3 \cos^2 2A + 2\sqrt{3} \cos 2A \cos 2B + \cos^2 2B = 4 \]

### Step 5: Combine the equations
Now we can add these two equations together. The sine and cosine identities will help us simplify:
Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 3(\sin^2 2A + \cos^2 2A) + (\sin^2 2B + \cos^2 2B) = 4 + 2\sqrt{3} \cos 2A \cos 2B \]
This simplifies to:
\[ 3 + 1 = 4 + 2\sqrt{3} \cos 2A \cos 2B \]
Thus we get:
\[ 4 = 4 + 2\sqrt{3} \cos 2A \cos 2B \]
This implies:
\[ 2\sqrt{3} \cos 2A \cos 2B = 0 \]
Thus, either \( \cos 2A = 0 \) or \( \cos 2B = 0 \).

### Step 6: Solve for \( A \) and \( B \)
1. If \( \cos 2A = 0 \):
\[ 2A = \frac{\pi}{2} + n\pi \implies A = \frac{\pi}{4} + \frac{n\pi}{2}, \quad n \in \mathbb{Z} \]

2. If \( \cos 2B = 0 \):
\[ 2B = \frac{\pi}{2} + m\pi \implies B = \frac{\pi}{4} + \frac{m\pi}{2}, \quad m \in \mathbb{Z} \]

### Final Result
Thus, the general solutions for \( A \) and \( B \) are:
\[ A = \frac{\pi}{4} + \frac{n\pi}{2}, \quad B = \frac{\pi}{4} + \frac{m\pi}{2}, \quad n, m \in \mathbb{Z} \]