JEE Main 2016 (8)

$\begin{aligned} \sin^2 x - 2 \sin x - 1 & = 0 \\ \Rightarrow \sin x & = \frac{2 \pm \sqrt{8}}{2} \\ & = 1 - \sqrt{2} \quad \text{since } \sin x \le 1 \end{aligned}$

Since $1-\sqrt{2} < 0$ , $\sin^{-1} (1-\sqrt{2}) < 0$ and let $\sin^{-1} (1-\sqrt{2}) = - \alpha$ , where $\alpha > 0$ .

Then, the first four positive values of $x = \begin{cases} \pi + \alpha \\ 2\pi - \alpha \\ 3\pi + \alpha \\ 4\pi - \alpha \end{cases}$ . Therefore the smallest $n=4$ and $n-1=\boxed{3}$ .

I graphed the function.