$$
\begin{aligned}
& f(x)=\cos ^2 x+\cos ^2 2 x+\cos ^2 3 x=1 \\
& \Rightarrow \cos ^2 x+\left(2 \cos ^2 x-1\right)^2+\left(4 \cos ^3 x-3 \cos x\right)^2=1 \\
& \Rightarrow \cos ^2 x+4 \cos ^4 x+1-4 \cos ^2 x+16 \cos ^6 x \\
& +9 \cos ^2 x-24 \cos ^4 x=1 \\
& \Rightarrow \quad 6 \cos ^2 x-20 \cos ^4 x+16 \cos ^6 x=0 \\
& \Rightarrow \quad 2 \cos ^2 x\left(3-10 \cos ^2 x+8 \cos ^4 x\right)=0 \\
& \Rightarrow \quad \cos ^2 x=0 \text { or } 8 \cos ^4 x-10 \cos ^2 x+3=0 \\
& \Rightarrow \quad \cos ^2 x=0 \text { or }\left(4 \cos ^2 x-3\right)\left(2 \cos ^2 x-1\right)=0 \\
& \Rightarrow \quad \cos ^2 x=0 \text { or } \cos x=\frac{3}{4} \text { or } \cos ^2 x=\frac{1}{2} \\
& \Rightarrow \quad \cos ^2 x=0 \text { or } \cos x=\frac{\sqrt{3}}{2} \text { or } \cos x=\frac{1}{\sqrt{2}} \\
& \Rightarrow \quad \cos ^2 x=(2 n+1) \frac{\pi}{2}, \\
& x=2 n \pi \pm \frac{\pi}{6} \\
&
\end{aligned}
$$

or
$$
\begin{array}{ll}
\text { or } & x=2 n \pi+\frac{\pi}{4} \\
\therefore & x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{\pi}{4}, \frac{\pi}{6}, \frac{7 \pi}{4}, \frac{11 \pi}{6}
\end{array}
$$
Hence, there are six values of $x$ for which $x \in[0,2 \pi]$.