答案：B
$\begin{array}{rcl}
\cos^2 x – \sin^2 x & = & 1 \\
\cos^2 x -(1-\cos^2 x) & = & 1 \\
2\cos^2 x- 1 & = & 1 \\
2\cos^2 x – 2& = & 0 \\
2(\cos^2 x -1) & = & 0 \\
(\cos x +1)(\cos x -1) & = & 0
\end{array}$
$\begin{array}{rcl}
\cos^2 x – \sin^2 x & = & 1 \\
\cos^2 x -(1-\cos^2 x) & = & 1 \\
2\cos^2 x- 1 & = & 1 \\
2\cos^2 x – 2& = & 0 \\
2(\cos^2 x -1) & = & 0 \\
(\cos x +1)(\cos x -1) & = & 0
\end{array}$
所以，$\cos x =-1$ 或 $\cos x =1$。對於 $\cos x = -1$，$x=180^\circ$。對於 $\cos x = 1$，$x=0^\circ$ 或 $x = 360^\circ$。
由此，該方程有 $3$ 個根。
