## Task A-3.1. (4 points)

Solve the equation $\quad \sin x \cdot \cos 2 x \cdot \cos 4 x=1$.
## First solution.

Since $\sin x, \cos 2 x$ and $\cos 4 x$ are numbers from the interval $[-1,1]$, we see that all three numbers must be 1 or -1.

1. case.

Let $\sin x=1$. Then $\cos x=0, \cos 2 x=\cos ^{2} x-\sin ^{2} x=-1, \sin 2 x=0$ and $\cos 4 x=\cos ^{2} 2 x-\sin ^{2} 2 x=1$. However, in this case, $\sin x \cdot \cos 2 x \cdot \cos 4 x=-1$, so there is no solution in this case.

2. case.

Let $\sin x=-1$. Then $\cos x=0, \cos 2 x=\cos ^{2} x-\sin ^{2} x=-1, \sin 2 x=0$ and $\cos 4 x=\cos ^{2} 2 x-\sin ^{2} 2 x=1$. In this case, $\sin x \cdot \cos 2 x \cdot \cos 4 x=-1 \cdot(-1) \cdot 1=1$, so the given equation is satisfied by all $x$ for which $\sin x=-1$.

The solution to the given equation are all numbers of the form $x=\frac{3 \pi}{2}+2 k \pi$ where $k$ is an integer.