Question

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

Answer 1

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
\[\theta = \frac{7\pi}{6},\frac{11\pi}{6},\frac{3\pi}{2}\]