Question

If ${\sin }^{2}x-2\sin x-1=0$ has exactly four different solutions in $x\in [0,n\pi ]$, then minimum value of $n$ can be $(n\in N)$

Answer 1

We have, ${\sin }^{2}x-2\sin x-1=0$

${(\sin x-1)}^2=2\Rightarrow \sin x-1=\pm \sqrt{2}$

$\Rightarrow \sin x=1-\sqrt{2}$ as $\sin x\ngtr 1$

There are $2$ solutions in $[0,2\pi ]$ and two more in $[2\pi ,4\pi ]$.

Thus, minimum value of $n$ is $4$.