Solve the following equation:

$$
\sin ^{4} x+\sin ^{4} 2 x+\sin ^{4} 3 x=\cos ^{4} x+\cos ^{4} 2 x+\cos ^{4} 3 x
$$
If we reduce equation (1) to zero and perform similar transformations as in the following pattern:

$$
\cos ^{4} \alpha-\sin ^{4} \alpha=\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)=\cos 2 \alpha
$$

we obtain the following equation:

$$
\cos 2 x+\cos 4 x+\cos 6 x=0
$$

Using known identities,

$$
\begin{aligned}
& \cos 2 x=\cos (4 x-2 x)=\cos 4 x \cdot \cos 2 x+\sin 4 x \cdot \sin 2 x, \quad \text { and } \\
& \cos 6 x=\cos (4 x+2 x)=\cos 4 x \cdot \cos 2 x-\sin 4 x \cdot \sin 2 x
\end{aligned}
$$

Using these, our equation can be written as:

$$
\cos 4 x \cdot(2 \cos 2 x+1)=0
$$

This holds if and only if either $\cos 4 x=0$ or $2 \cos 2 x+1=0$. In the first case, $\cos 4 x=0$ if $4 x=\frac{\pi}{2}+k \pi$, that is,

$$
x_{1}=\frac{\pi}{8}+k \frac{\pi}{4}=(2 k+1) \frac{\pi}{8}
$$

In the second case, $\cos 2 x=-\frac{1}{2}$. Then $2 x= \pm \frac{2 \pi}{3}+2 k \pi$, so

$$
x_{2}= \pm \frac{\pi}{3}+k \pi=(3 k \pm 1) \frac{\pi}{3}
$$

($k$ represents any integer in both cases.) Since our transformations were reversible, these and only these values of $x$ satisfy equation (1).

Gyula Mészáros (Budapest, Leövey K. Gymnasium, 4th grade)