2. 2. 15 * Solve the equation: $\cos ^{2} x+\cos ^{2} 2 x+\cos ^{2} 3 x=1$.
Applying the double-angle formula of the cosine function to the first two terms on the left side of equation (1), we can reduce the power, yielding
$$
\cos 2 x+\cos 4 x+2 \cos ^{2} 3 x=0 .
$$

We observe that, by transforming the sum and difference of the first two terms on the left side of equation (2) into a product, we can then factor out the common factor $2 \cos 3 x \cos x + 2 \cos ^{2} 3 x=0 \Rightarrow \cos 3 x(\cos x+\cos 3 x)=0 \Rightarrow \cos x \cos 2 x \cos 3 x=0$.
Thus, we obtain three sets of solutions for the original equation: $x_{1}=(2 k+1) \frac{\pi}{2}, x_{2}=(2 k+1) \frac{\pi}{4}, x_{3}=(2 k+1) \frac{\pi}{6}, k \in \mathbf{Z}$.