Ans: B
$\begin{array}{rcl}
6\cos^2 x & = & \cos x + 5 \\
6\cos^2 – \cos x – 5 & = & 0 \\
(6\cos x +5)(\cos x – 1) & = & 0
\end{array}$
$\begin{array}{rcl}
6\cos^2 x & = & \cos x + 5 \\
6\cos^2 – \cos x – 5 & = & 0 \\
(6\cos x +5)(\cos x – 1) & = & 0
\end{array}$
Therefore, $\cos x = \dfrac{-5}{6}$ or $\cos x = 1$.
For $\cos x = \dfrac{-5}{6}$, we have
$\begin{array}{rcl}
\cos x & = & \dfrac{-5}{6} \\
x & = & 146.442\ 690\ 2^\circ \text{ or } 213.557\ 309\ 8^\circ
\end{array}$
For $\cos x = 1$, we have $x = 270^\circ$.
Therefore, there are $3$ roots.
