2008-II-45

Ans: B
Let $y = \sin \theta$. Then the equation becomes

$\begin{array}{rcl}
3y^2 + 2y -1 & = & 0 \\
(3y-1)(y+1) & = & 0 \\
y=\dfrac{1}{3} & \text{or} & y = -1 \\
\sin \theta = \dfrac{1}{3} & \text{or} & \sin \theta = -1
\end{array}$

For $\sin \theta = \dfrac{1}{3}$, we have

$\begin{array}{rcl}
\sin \theta & = & \dfrac{1}{3} \\
\theta & = & 19.471~220~63^\circ \\
& \text{ or } & 180^\circ – 19.471~220~63^\circ \\
& = & 160.528~779~4^\circ
\end{array}$

For $\sin \theta = -1$, we have

$\begin{array}{rcl}
\sin \theta & = & -1 \\
\theta & = & 270^\circ
\end{array}$

Therefore, there are $3$ roots.