A5 (4-4, Romania) Solve the equation $\cos ^{2} x+\cos ^{2} 2 x+\cos ^{2} 3 x=1$.
Solve: Substituting $\cos ^{2} x=\frac{1}{2}(1+\cos 2 x), \cos ^{2} 3 x=\frac{1}{2}(1+\cos 6 x)$ into the given equation, we get
$$
\cos 2 x+\cos 6 x+2 \cos ^{2} 2 x=0 .
$$

Applying the sum-to-product formula to the first two terms, we get
$$
\cos 2 x(\cos 4 x+\cos 2 x)=0 .
$$

Applying the sum-to-product formula again to the terms in the parentheses, we get
$$
\cos x \cos 2 x \cos 3 x=0 .
$$

Thus, we obtain
$$
\begin{array}{l}
x_{1}=k \pi+\frac{\pi}{2}, \\
x_{2}=\frac{k \pi}{2}+\frac{\pi}{4}, \\
x_{3}=\frac{k \pi}{3}+\frac{\pi}{6} .
\end{array}
$$

Here $k$ runs through all integers, and it is easy to see that the first solution $x_{1}$ is included in the third solution $x_{3}$, so it can be discarded.