Problem 2. For any positive integer $n$, let the functions $f_{n}: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f_{n+1}(x)=f_{1}\left(f_{n}(x)\right)$, where $f_{1}(x)=3 x-4 x^{3}$. Solve the equation $f_{n}(x)=0$.
Solution. First, we prove that $|x|>1 \Rightarrow\left|f_{n}(x)\right|>1$ holds for every positive integer $n$. It suffices to demonstrate the validity of this implication for $n=1$. But, by assuming $|x|>1$, it readily follows that $\left|f_{1}(x)\right|=|x|\left|3-4 x^{2}\right| \geq\left|3-4 x^{2}\right|>1$, which completes the demonstration. We conclude that every solution of the equation $f_{n}(x)=0$ lies in the closed interval $[-1,1]$. For an arbitrary such $x$, set $x=\sin t$ where $t=\arcsin x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. We clearly have $f_{1}(\sin t)=\sin 3 t$, which gives

$$
f_{n}(x)=\sin 3^{n} t=\sin \left(3^{n} \arcsin x\right)
$$

Thus, $f_{n}(x)=0$ if and only if $\sin \left(3^{n} \arcsin x\right)=0$, i.e. only when $3^{n} \arcsin x=k \pi$ for some $k \in \mathbb{Z}$. Therefore, the solutions of the equation $f_{n}(x)=0$ are given by

$$
x=\sin \frac{k \pi}{3^{n}}
$$

where $k$ acquires every integer value from $\frac{1-3^{n}}{2}$ up to $\frac{3^{n}-1}{2}$.