Question

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

• A. 5
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (A), (C)

The correct options are
A 5
C 4

$si{n}^2\theta -2sin\theta -1=0$
$sin\theta =1\pm \sqrt{2}$, but $-1\le sin\theta \le 1$
so the only solution is,
$sin\theta =-(\sqrt{2}-1)=sin(-\alpha )$, (suppose)
therefore general solution is,
$\theta =n\pi -(-1{)}^n\alpha$
in $[0,n\pi ]$
$\theta =\pi +\alpha ,2\pi -\alpha ,3\pi +\alpha ,4\pi -\alpha ,5\pi +\alpha ,....$
Hence for 4 distinct solution least value of $nϵN$ is 4 and greatest value is 5.
( it will be easy to see it by drawing graph of $sinx=1-\sqrt{2}$.