Let `x` and y, `(0 lt x, y lt pi/2)` satisfy `3sin^(2)x + 2sin^(2)y=1` and `3sin2x = 2sin2y`, then the value of `232/pi (x+2y)` is……………

Explanation

To solve the problem, we need to find the value of \( \frac{232}{\pi}(x + 2y) \) given the equations:

1. \( 3\sin^2 x + 2\sin^2 y = 1 \)
2. \( 3\sin 2x = 2\sin 2y \)

Let's break down the solution step by step.

### Step 1: Rearranging the first equation
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