anonymous 5 years ago Solve the given equation and indicate the number of values of θ sach that 0≤θ<2π,that satisfy the iquation ; 2sin^2θ-sinθ-1=0

Take $\sin\theta=x$ given equation $2x^2-x-1=0$ solving above equation, we get x=-0.5 or x=1 or we can say $\sin\theta = -0.5$ or $\sin\theta = 1$ so we can say the values of theta for which this is true is $\theta = \frac{7\pi}{6},\frac{11\pi}{6},\frac{3\pi}{2}$