If `alpha` and `beta` are `+` ive acute angle satisfying the equation `3 sin^(2) alpha +2 sin^(2) beta =1` and `3 sin 2 alpha -2 sin 2 beta =0`, then `alpha +2 beta=`

Options

A. `pi//2`
B. `2pi//3`
C. `pi`
D. none

Answer

To solve the problem, we have the following equations:

1. \( 3 \sin^2 \alpha + 2 \sin^2 \beta = 1 \)
2. \( 3 \sin 2\alpha - 2 \sin 2\beta = 0 \)

We need to find \( \alpha + 2\beta \).

### Step 1: Rearranging the First Equation
From the first equation, we can express \( \sin^2 \alpha \) in terms of \( \sin^2 \beta \):

\[ 3 \sin^2 \alpha = 1 - 2 \sin^2 \beta \]

### Step 2: Expressing \( \sin^2 \alpha \)
Now, we can isolate \( \sin^2 \alpha \):

\[ \sin^2 \alpha = \frac{1 - 2 \sin^2 \beta}{3} \]

### Step 3: Using the Second Equation
From the second equation, we can express \( \sin 2\alpha \) in terms of \( \sin 2\beta \):

\[ 3 \sin 2\alpha = 2 \sin 2\beta \]

### Step 4: Expressing \( \sin 2\alpha \)
Using the double angle formula \( \sin 2\theta = 2 \sin \theta \cos \theta \):

\[ 3 \cdot 2 \sin \alpha \cos \alpha = 2 \cdot 2 \sin \beta \cos \beta \]

This simplifies to:

\[ 3 \sin \alpha \cos \alpha = 2 \sin \beta \cos \beta \]

### Step 5: Equating the Two Equations
Now we have two equations:

1. \( \sin^2 \alpha = \frac{1 - 2 \sin^2 \beta}{3} \)
2. \( 3 \sin \alpha \cos \alpha = 2 \sin \beta \cos \beta \)

We can express \( \sin \alpha \) and \( \sin \beta \) in terms of each other using these equations.

### Step 6: Substituting Values
Substituting the value of \( \sin^2 \alpha \) into the second equation will allow us to find a relationship between \( \alpha \) and \( \beta \).

### Step 7: Using the Cosine Identity
From the equations, we can derive:

\[ \cos(2\beta) \cos(\alpha) - \sin(2\beta) \sin(\alpha) = 0 \]

This can be rewritten using the cosine of a sum:

\[ \cos(\alpha + 2\beta) = 0 \]

### Step 8: Finding the Angle
From \( \cos(\alpha + 2\beta) = 0 \), we know that:

\[ \alpha + 2\beta = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

Since \( \alpha \) and \( \beta \) are acute angles, we take \( n = 0 \):

\[ \alpha + 2\beta = \frac{\pi}{2} \]

### Final Answer
Thus, the value of \( \alpha + 2\beta \) is:

\[ \alpha + 2\beta = \frac{\pi}{2} \]