If `Aa n dB` are acute positive angles satisfying the equations `3sin^2A+2sin^2B=1a n d3sin2A-2sin2B=0,t h e nA+2B` is equal to `pi` (b) `pi/2` (c) `pi/4` (d) `pi/6`

Options

A. `pi/4`
B. `pi/2`
C. `(3pi)/(4)`
D. `(2pi)/(3)`

Answer

To solve the problem, we need to analyze the given equations step by step.

### Step 1: Analyze the first equation
The first equation given is:
\[ 3\sin^2 A + 2\sin^2 B = 1 \]

Rearranging this equation gives us:
\[ 3\sin^2 A = 1 - 2\sin^2 B \]

### Step 2: Use the identity for cosine
We can express \( 1 - 2\sin^2 B \) using the cosine double angle identity:
\[ 1 - 2\sin^2 B = \cos 2B \]
Thus, we rewrite the equation as:
\[ 3\sin^2 A = \cos 2B \]

### Step 3: Analyze the second equation
The second equation given is:
\[ 3\sin 2A - 2\sin 2B = 0 \]

Rearranging this gives us:
\[ 3\sin 2A = 2\sin 2B \]
Dividing both sides by 2:
\[ \frac{3}{2}\sin 2A = \sin 2B \]

### Step 4: Substitute \( \sin 2B \)
From the previous step, we have:
\[ \sin 2B = \frac{3}{2}\sin 2A \]

### Step 5: Substitute \( \sin 2B \) into the cosine equation
Using the identity \( \sin 2B = 2\sin B \cos B \), we can express \( \sin 2B \) in terms of \( A \):
\[ 2\sin B \cos B = \frac{3}{2}(2\sin A \cos A) \]
This simplifies to:
\[ 2\sin B \cos B = 3\sin A \cos A \]

### Step 6: Solve for \( A + 2B \)
From the rearranged equations, we can derive a relationship between \( A \) and \( B \). We know that:
\[ \cos 2B = 3\sin^2 A \]
And from the second equation, we can derive:
\[ \sin 2B = 3\sin A \cos A \]

Using the cosine and sine relationships, we can find that:
\[ A + 2B = \frac{\pi}{2} \]

### Conclusion
Thus, the value of \( A + 2B \) is:
\[ \boxed{\frac{\pi}{2}} \]