Question

If ${\sin }^2\theta -2\sin \theta -1=0$ is to be satisfied for exactly 4 distinct values of $\theta ϵ[0,n\pi ],nϵN$, then the least value of n is

• A. 2
• B. 6
• C. 4
• D. 1

Answer 1

Answer: (C)

4

${\sin }^2\theta -2\sin \theta -1=0$

$\sin \theta =1\pm \sqrt{2}$, 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 $[0,n\pi ]$

$\theta =\pi +\alpha ,2\pi -\alpha ,3\pi +\alpha ,4\pi -\alpha ,$

Hence for 4 distinct solution least value of $n\in N$ is 4.