Question

If ${\sin }^{2}x-2\sin x-1=0$ has exactly four different solutions in the interval $[0,n\pi ]$ , then possible value(s) of $n$ is/are :

• A. $5$
• B. $3$
• C. $4$
• D. $6$

Answer 1

Answer: (A), (C)

The correct options are
A $5$
C $4$

${\sin }^{2}x-2\sin x-1=0\Rightarrow (\sin x-1{)}^2=2\Rightarrow \sin x-1=\pm \sqrt{2}\Rightarrow \sin x=1-\sqrt{2}(\because \sin x\le 1)$

We know that $\sin x$ is positive in $[0,\pi ]$ and negative in $[\pi ,2\pi ].$
$\Rightarrow \sin x=1-\sqrt{2}<0$ is possible only in $3\text{rd}$ and $4\text{th}$ quadrant.
So, there are only $2$ solutions in $[0,2\pi ].$
Since, period of $\sin x$ is $2\pi ,$ hence we get two more in $[2\pi ,4\pi ].$


$\therefore n=4,5$