Ans: A
$\begin{array}{rcl}
3\cos^2 x -4\cos x +1 & = & 0 \\
(3\cos x -1)(\cos x -1) & = & 0
\end{array}$
$\begin{array}{rcl}
3\cos^2 x -4\cos x +1 & = & 0 \\
(3\cos x -1)(\cos x -1) & = & 0
\end{array}$
Therefore, $\cos x = \dfrac{1}{3}$ or $\cos x = 1$.
For $\cos x =\dfrac{1}{3}$, we have
$\begin{array}{rcl}
\cos x & = & \dfrac{1}{3} \\
x & = & 70.5^\circ~\text{or}~289^\circ
\end{array}$
For $\cos x =1$, we have
$\begin{array}{rcl}
\cos x & = & 1 \\
x & = & 0^\circ~\text{or}~360^\circ
\end{array}$
Since $0^\circ < x < 360^\circ$, then $x=70.5^\circ$ or $x=289^\circ$. Hence, there are two roots.
