Ans: B
$\begin{array}{rcl}
\cos^2 x – \sin x & = & 1 \\
(1-\sin^2 x) – \sin x -1 & = & 0 \\
-\sin^2 x -\sin x & = & 0 \\
\sin x( \sin x – 1) & = & 0
\end{array}$
$\begin{array}{rcl}
\cos^2 x – \sin x & = & 1 \\
(1-\sin^2 x) – \sin x -1 & = & 0 \\
-\sin^2 x -\sin x & = & 0 \\
\sin x( \sin x – 1) & = & 0
\end{array}$
Therefore, $\sin x =0$ or $\sin x = 1$.
Hence, we have $x=0^\circ$, $x = 180^\circ$ or $x=90^\circ$.
Therefore, there are $3$ roots.
