答案:B
$\begin{array}{rcl}
6\cos^2 x & = & \cos x + 5 \\
6\cos^2 – \cos x – 5 & = & 0 \\
(6\cos x +5)(\cos x – 1) & = & 0
\end{array}$
$\begin{array}{rcl}
6\cos^2 x & = & \cos x + 5 \\
6\cos^2 – \cos x – 5 & = & 0 \\
(6\cos x +5)(\cos x – 1) & = & 0
\end{array}$
所以,$\cos x = \dfrac{-5}{6}$ 或 $\cos x = 1$。
對於 $\cos x = \dfrac{-5}{6}$,可得
$\begin{array}{rcl}
\cos x & = & \dfrac{-5}{6} \\
x & = & 146.442\ 690\ 2^\circ \text{ 或 } 213.557\ 309\ 8^\circ
\end{array}$
對於 $\cos x = 1$,可得 $x = 270^\circ$。
所以,有 $3$ 個根。