Task B-3.1. Solve the equation $\cos ^{2} x+\cos ^{2}(2 x)+\cos ^{2}(3 x)=1$.
## Solution.

Apply the following identities in the given equation

$$
\cos ^{2} x=\frac{1+\cos 2 x}{2}, \cos ^{2} 2 x=\frac{1+\cos 4 x}{2}
$$

It follows that

$$
\begin{aligned}
& \frac{1+\cos 2 x}{2}+\frac{1+\cos 4 x}{2}+\cos ^{2} 3 x=1 \\
& 1+\cos 2 x+1+\cos 4 x+2 \cos ^{2} 3 x=2 \\
& \cos 2 x+\cos 4 x+2 \cos ^{2} 3 x=0 \\
& 2 \cos \frac{2 x+4 x}{2} \cos \frac{2 x-4 x}{2}+2 \cos ^{2} 3 x=0 \\
& 2 \cos 3 x \cdot \cos x+2 \cos ^{2} 3 x=0 \\
& 2 \cos 3 x \cdot(\cos x+\cos 3 x)=0 \\
& 4 \cos 3 x \cdot \cos 2 x \cdot \cos x=0 \\
& \cos 3 x=0 \Rightarrow x=\frac{\pi}{6}+\frac{k \pi}{3}, k \in \mathbb{Z} \\
& \cos 2 x=0 \Rightarrow x=\frac{\pi}{4}+\frac{k \pi}{2}, k \in \mathbb{Z} \\
& \cos x=0 \Rightarrow x=\frac{\pi}{2}+k \pi, k \in \mathbb{Z}
\end{aligned}
$$