Number of non negative integral values of $n,n\le 10$ so that a root of the equation $n^2\sin^2 x-2\sin x-(2n+1)=0$ lies in the interval $[0,\pi/2]$ Find the number of non negative integral values of $n,n\le 10$ so that a root of the equation $n^2\sin^2 x-2\sin x-(2n+1)=0$ lies in the interval $[0,\pi/2]$

As $x\in[0,\pi/2]$ so $\sin x\in[0,1]$ and letting $\sin x=t$,we get $n^2t^2-2t-2n-1=0$ I am stuck here.

$n^2 t^2 -2t - 2n -1=0$ is a quadratic in $t$. Solving it for $t$, we get:
$$
\begin{align} 
t &= \frac{2 \pm \sqrt{4-4n^2(-2n-1)}}{2n^2}
\\ &= \frac{1 \pm \sqrt{1+n^2(2n+1)}}{n^2}
\end{align}
$$
It can easily be checked that $n=0$ does not lead to $t \in [0,1]$ so we may assume that $n \neq 0$.
Now for $t \in [0,1]$ we must have:
$$
0 \leq \frac{1 \pm \sqrt{1+n^2(2n+1)}}{n^2} \leq 1
\\ \implies 0 \leq 1 \pm \sqrt{1+n^2(2n+1)} \leq n^2
\\ \implies -1 \leq \pm \sqrt{1+n^2(2n+1)} \leq n^2 -1
\\ \implies 1 \leq 1+n^2(2n+1) \leq n^4 -2n^2 +1
\\ \implies 0 \leq n^2(2n+1) \leq n^4 -2n^2
\\ \implies 0 \leq 2n+1 \leq n^2 -2
$$
For $0 \leq n$ we automatically have $0 \leq 2n+1$. Solving the quadratic inequality, we get $3 \leq n$.
