11.2. Find all solutions of the equation: $\cos ^{2} x+\cos ^{2} 2 x+\cos ^{2} 3 x=1$.
Answer. $x= \pm \frac{\pi}{2}+2 k \pi, \pm \frac{\pi}{4}+\frac{l \pi}{2}, \pm \frac{\pi}{6}$

Solution. One can, acting straightforwardly, replace

after transformations we get: $8 \cos ^{6} x-10 \cos ^{4} x+3 \cos ^{2} x=0$, from which $\cos x=0, \pm \frac{1}{\sqrt{2}}, \pm \frac{\sqrt{3}}{2}$. Therefore, $x= \pm \frac{\pi}{2}+2 k \pi, \pm \frac{\pi}{4}+\frac{l}{2}, \pm \frac{\pi}{6}+m \pi$

One can solve it differently, by replacing the squares of cosines with cosines of double angles, obtaining $\cos 2 x+\cos 4 x+\cos 6 x=-1$, then folding the sum of the first and third terms into a double product of cosines, expressing everything in terms of $\cos 2 x$, obtaining the equation $2 \cos ^{3} 2 x+\cos ^{2} 2 x-\cos 2 x=0$ with solutions $\cos 2 x=0,-1, \frac{1}{2}$.

Grading criteria. Finding the values $\cos x=0, \pm \frac{1}{\sqrt{2}}, \pm \frac{\sqrt{3}}{2}$ or $\cos 2 x=0,-1, \frac{1}{2}: 4$ points. Correctly listing all series of solutions: 3 points. Losing part of the solutions: deduct 2-3 points.