Let θ ∈ [-2π, 2π] satisfying 2cos^2θ - sinθ - 1 = 0. Then the number of solutions of equation is

Question

Let \(\theta \in[-2 \pi, 2 \pi]\) satisfying \(2 \cos ^{2} \theta-\sin \theta-1=0.\)

Then the number of solutions of equation is

(1) 2

(2) 4

(3) 6

(4) 8

Answer 1

Correct option is: (3) 6

\(2 \cos ^{2} \theta-\sin \theta-1=0\)

\(2\left(1-\sin ^{2} \theta\right)-\sin \theta-1=0\)

\(\Rightarrow-2 \sin ^{2} \theta-\sin \theta+1=0\)

\(\Rightarrow 2 \sin ^{2} \theta+\sin \theta-1=0\)

\(\Rightarrow 2 \sin ^{2} \theta+2 \sin \theta-\sin \theta-1=0\)    

\(\Rightarrow 2 \sin \theta(\sin \theta+1)-1(\sin \theta+1)=0\)

\(\Rightarrow(2 \sin \theta-1)(\sin \theta+1)=0\)

\(\Rightarrow \sin \theta=\frac{1}{2},-1\)