Ans: B
$\begin{array}{rcl}
\cos^2 x – \sin^2 x & = & 1 \\
\cos^2 x -(1-\cos^2 x) & = & 1 \\
2\cos^2 x- 1 & = & 1 \\
2\cos^2 x – 2& = & 0 \\
2(\cos^2 x -1) & = & 0 \\
(\cos x +1)(\cos x -1) & = & 0
\end{array}$
$\begin{array}{rcl}
\cos^2 x – \sin^2 x & = & 1 \\
\cos^2 x -(1-\cos^2 x) & = & 1 \\
2\cos^2 x- 1 & = & 1 \\
2\cos^2 x – 2& = & 0 \\
2(\cos^2 x -1) & = & 0 \\
(\cos x +1)(\cos x -1) & = & 0
\end{array}$
Therefore, $\cos x =-1$ or $\cos x =1$. For $\cos x = -1$, $x=180^\circ$. For $\cos x = 1$, $x=0^\circ$ or $x = 360^\circ$.
Hence, the equation has $3$ roots.
