Solve the following equation:

$(\sin 3 x+\sqrt{3} \cos 3 x)^{2}-2 \cos 14 x=2$.
Let's transform our equation and divide both sides by 4:

$$
\left(\frac{1}{2} \sin 3 x+\frac{\sqrt{3}}{2} \cos 3 x\right)^{2}=\frac{1+\cos 14 x}{2}
$$

Since $\frac{1}{2}=\sin \frac{\pi}{6}$ and $\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}$, based on the identities $\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta$ and $\cos ^{2} \alpha=(1+\cos 2 \alpha) / 2$, it follows that

$$
\frac{1+\cos \left(6 x-\frac{\pi}{3}\right)}{2}=\frac{1+\cos 14 x}{2}
$$

which simplifies to

$$
\cos \left(6 x-\frac{\pi}{3}\right)=\cos 14 x
$$

From this, we get

$$
6 x-\frac{\pi}{3}= \pm 14 x+2 k \pi, \quad k=0, \pm 1, \pm 2, \ldots
$$

and

$$
x_{1}=\frac{\pi}{60}+k \frac{\pi}{10}, \quad x_{2}=-\frac{\pi}{24}-k \frac{\pi}{4}
$$

Since we only performed equivalent transformations during the solution process, the obtained values are indeed roots of the equation.

Révész Sz. György (Budapest, I. István Gymn., IV. o. t.)