Question:

<p><strong>Step 1:</strong> Using Shridharacharya's formula on $n^2\sin^2 x + 2\sin x - (2n+1) = 0$:</p><p>$\sin x = \frac{-2 \pm \sqrt{4 + 4n^2(2n+1)}}{2n^2} = \frac{-1 \pm \sqrt{2n^3 + n^2 + 1}}{n^2}$</p><p><strong>Step 2:</strong> Since $x \in [0, \frac{\pi}{2}]$, we have $0 \le \sin x \le 1$.</p><p><strong>Step 3:</strong> Taking the positive root: $\sin x = \frac{-1 + \sqrt{2n^3 + n^2 + 1}}{n^2}$</p><p><strong>Step 4:</strong> For $0 \le \sin x \le 1$:</p><p>$0 \le 1 + \sqrt{2n^3 + n^2 + 1} \le n^2$</p><p>$\sqrt{2n^3 + n^2 + 1} \le n^2 - 1$</p><p><strong>Step 5:</strong> Squaring: $2n^3 + n^2 + 1 \le n^4 - 2n^2 + 1$</p><p>$n^4 - 2n^3 - 3n^2 \ge 0$</p><p>$n^2(n^2 - 2n - 3) \ge 0$</p><p>$(n-3)(n+1) \ge 0$</p><p>$n \ge 3$</p><p>∴ Minimum positive integer value of $n$ is <strong>3</strong>.</p>