Ans: D
$\begin{array}{rcl}
7\sin^2 x & = & \sin x \\
7\sin^2 x – \sin x & = & 0 \\
\sin x(7\sin x -1) & = & 0
\end{array}$
$\begin{array}{rcl}
7\sin^2 x & = & \sin x \\
7\sin^2 x – \sin x & = & 0 \\
\sin x(7\sin x -1) & = & 0
\end{array}$
Therefore, $\sin x=0$ or $\sin x =\dfrac{1}{7}$.
For $\sin x =0$, we have $x = 0^\circ$, $180^\circ$ or $360^\circ$.
For $\sin x = \dfrac{1}{7}$, we have $x=8.213\ 210\ 702^\circ$ or $171.786\ 789\ 3^\circ$.
Therefore, there are $5$ roots.
