Question

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

• A. $5$
• B. $3$
• C. $4$
• D. $6$

Answer 1

Answer: (A), (C)

$5$
$4$

${\sin }^{2}x-2\sin x-1=0$
$\Rightarrow {(\sin x-1)}^2=2$
$\Rightarrow \sin x-1=\pm \sqrt{2}$
$\Rightarrow \sin x=1-\sqrt{2}$ as $\sin x\ngtr 1$
$\sin x<0$, Hence $x$ can be in the third or fourth quadrant.
There are two solutions in $[0,2\pi ]$ and two more in $[2\pi ,4\pi ]$
If we consider $n=3$ , there will be $2$ solutions
If we consider $n=4$ , there will be $4$ solutions
If we consider $n=5$, there will be $4$ solutions.
If we consider $n=6$, there will be $6$ solutions
Hence, options A,C are correct