Question

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

• A. 2
• B. 6
• C. 4
• D. 15

Answer 1

The correct option is C 4

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

${\sin }^2\theta -2\sin \theta -({\sin }^2\theta +{\cos }^2\theta )=0$

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

$2\cos \theta \cdot \sin \theta -2\sin \theta =0$

$2\sin \theta (\cos \theta -1)=0$

$\sin \theta =0$ or $\cos \theta =1$

$\theta =0,\pi ,2\pi ,3\pi ....$ $\theta =0,2\pi ,4\pi .....$

$\therefore \theta \in [0,4\pi ]$

$\therefore$ Least value of $n$ is $4$.