答案：A
$\begin{array}{rcl}
3\cos^2 x -4\cos x +1 & = & 0 \\
(3\cos x -1)(\cos x -1) & = & 0
\end{array}$
$\begin{array}{rcl}
3\cos^2 x -4\cos x +1 & = & 0 \\
(3\cos x -1)(\cos x -1) & = & 0
\end{array}$
所以，$\cos x = \dfrac{1}{3}$ 或 $\cos x = 1$。
對於 $\cos x =\dfrac{1}{3}$，可得
$\begin{array}{rcl}
\cos x & = & \dfrac{1}{3} \\
x & = & 70.5^\circ~\text{or}~289^\circ
\end{array}$
對於 $\cos x =1$，可得
$\begin{array}{rcl}
\cos x & = & 1 \\
x & = & 0^\circ~\text{or}~360^\circ
\end{array}$
因為 $0^\circ < x < 360^\circ$，則 $x=70.5^\circ$ 或 $x=289^\circ$。由此，該方程有兩個根。
