答案：B
$\begin{array}{rcl}
5 \sin^2 \theta + \sin \theta – 4 & = & 0 \\
(5 \sin \theta -4)(\sin \theta +1) & = & 0 \\
\end{array}$
$\begin{array}{rcl}
5 \sin^2 \theta + \sin \theta – 4 & = & 0 \\
(5 \sin \theta -4)(\sin \theta +1) & = & 0 \\
\end{array}$
所以，$\sin \theta = \dfrac{4}{5}$ 或 $\sin \theta =-1$。
考慮 $\sin \theta = \dfrac{4}{5}$，$\theta = 53.1^\circ$ 或 $\theta = 127^\circ$。
考慮 $\sin \theta = -1$，$\theta = 270^\circ$。
所以，該方程有 $3$ 個根。
