If a root of the equation `n^(2)sin^(2)x-2sinx-(2n+1)=0` lies in`[0,pi//2]` the minimum positive integer value of `n` is

Explanation

To solve the equation \( n^2 \sin^2 x - 2 \sin x - (2n + 1) = 0 \) and find the minimum positive integer value of \( n \) such that a root lies in the interval \([0, \frac{\pi}{2}]\), we can follow these steps:

### Step 1: Identify the coefficients
The given equation can be rewritten in the standard quadratic form \( Ax^2 + Bx + C = 0 \), where:
- \( A = n^2 \)
- \( B = -2 \)
- \( C = -(2n + 1) \)

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