Croatian Mathematical Competitions

We have $|\sin x| \leq 1$, $|\cos 2x| \leq 1$, $|\cos 4x| \leq 1$. The product $\sin x \cdot \cos 2x \cdot \cos 4x$ can only be $1$ if each factor is $1$ or $-1$, and their product is $1$.

But $|\sin x| = 1$ only when $\sin x = 1$ or $\sin x = -1$.

Case 1: $\sin x = 1$
Then $x = \dfrac{\pi}{2} + 2k\pi$, $k \in \mathbb{Z}$.

At $x = \dfrac{\pi}{2}$:
$\cos 2x = \cos(\pi) = -1$
$\cos 4x = \cos(2\pi) = 1$
So $\sin x \cdot \cos 2x \cdot \cos 4x = 1 \cdot (-1) \cdot 1 = -1 \neq 1$

Case 2: $\sin x = -1$
Then $x = \dfrac{3\pi}{2} + 2k\pi$, $k \in \mathbb{Z}$.

At $x = \dfrac{3\pi}{2}$:
$\cos 2x = \cos(3\pi) = -1$
$\cos 4x = \cos(6\pi) = 1$
So $\sin x \cdot \cos 2x \cdot \cos 4x = (-1) \cdot (-1) \cdot 1 = 1$

Therefore, all solutions are $x = \dfrac{3\pi}{2} + 2k\pi$, $k \in \mathbb{Z}$.

Answer: $x = \dfrac{3\pi}{2} + 2k\pi$, $k \in \mathbb{Z}$.