1. Solve the equation in the interval $[0,2 \pi]$

$$
\cos ^{2} x+\cos ^{2} 2 x+\cos ^{2} 3 x=1
$$
1. By substituting the identities $\cos 2 x=2 \cos ^{2} x-1$ and $\cos 3 x=4 \cos ^{3} x-3 \cos x$ into the given equation, we obtain

$$
\cos ^{2} x+\left(2 \cos ^{2} x-1\right)^{2}+\left(4 \cos ^{3} x-3 \cos x\right)^{2}=1
$$

which, after simplification, gives $16 \cos ^{6} x-20 \cos ^{4} x+6 \cos ^{2} x=0$. By introducing the substitution $y=\cos ^{2} x$, the previous equation reduces to $16 y^{3}-20 y^{2}+6 y=0$,
i.e., $y\left(8 y^{2}-10 y+3\right)=0$, whose solutions are $y_{1}=0$ and $y_{2 / 3}=\frac{10 \pm \sqrt{10^{2}-4 \cdot 8 \cdot 3}}{16}=\frac{10 \pm \sqrt{4}}{16}$, i.e., $y_{2}=\frac{3}{4}$ and $y_{3}=\frac{1}{2}$. Reverting the substitution gives $\cos x \in\left\{0, \pm \frac{\sqrt{3}}{2}, \pm \frac{1}{2}\right\}$, which in the first case leads to the solution $x \in\left\{\frac{\pi}{2}, \frac{3 \pi}{2}\right\}$, in the second case $x \in\left\{\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right\}$, and in the third case $x \in\left\{\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}\right\}$. Therefore, the given equation has a total of 10 solutions on the interval $[0,2 \pi]$.