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We have $|\sin x|\le 1$, $|\cos 2x|\le 1$, $|\cos 4x|\le 1$. 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 $x=\frac{\pi }{2}+2k\pi$, $k\in \mathbb{Z}$.

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

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

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

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

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