1. Solve the equation: $\cos ^{2} x+\cos ^{2} 2 x=1$.
Answer: $\frac{\pi}{2}+\pi k, \pm \frac{\pi}{6}+\pi k$ (for integer $k$).

## Solution:

Since $\cos 2 x=2 \cos ^{2} x-1$, by substituting $y=\cos ^{2} x$, the equation reduces to $y+(2 y-1)^{2}=1$. Expanding and simplifying, we get $4 y^{2}-3 y=0$, from which $y=0$ or $y=\frac{3}{4}$.

The first of these values leads to $\cos ^{2} x=0 \Leftrightarrow \cos x=0 \Leftrightarrow x=\frac{\pi}{2}+\pi k$ (for integer $k$).

From the second value, we get $\cos ^{2} x=\frac{3}{4} \Leftrightarrow \cos x= \pm \frac{\sqrt{3}}{2} \Leftrightarrow x= \pm \frac{\pi}{6}+2 \pi k$ or $x= \pm \frac{2 \pi}{6}+2 \pi k$ (for integer $k$).

Note that the two series of solutions can be combined into $x= \pm \frac{\pi}{6}+\pi k$ (for integer $k$), but this is not particularly significant.

## Criteria:

Full solution found (all series) - 2 points, period lost - 1 point. Equation reduced to quadratic or biquadratic - plus 1 point.