The sequences of $\sin (a_n⋅x)$ Let $a_n$ be an increasing sequence of positive integers. Could it be that for any real number $x$ from the interval $[0;1]$ the sequence $\sin(a_n \cdot x)$ converges as $n→∞$?
due to poor knowledge of English, I incorrectly formulated his  question
 A: Well, this surely can happen for some sequences and some values of $x$. Say, $a_n=n!,\;x=2\pi e$ (or $2\pi(e-{8\over3})$, if you insist on $0<x\le1$); then $\lim\limits_{n\to\infty}\sin(a_n\cdot x)=0$.

See, $e=\sum\limits_{i=0}^\infty{1\over i!}$. When we multiply that by $n!$, the terms up to $1\over n!$ turn to integers, and what remains is $n!\left({1\over(n+1)!}+{1\over(n+2)!}+\dots\right)={1\over n+1}+{1\over(n+1)(n+2)}+\dots<{1\over n}$, so $|\sin(2\pi e\cdot n!)|<{2\pi\over n}\mathop{\longrightarrow}\limits_{n\to\infty}0$

If you have a sequence and want to find $x$ (besides the obvious $x=0$), that's not always possible.
If you have $x$ and want to find the sequence, it is always possible. But different values of $x$ would require different sequences; there is no "universal" sequence that would work for all (or "almost all") $x$.
A: At least the reverse holds: for every $x$, there exists some increasing sequence of integers $(a_n)$ (depending on $x$!) such that $\sin(a_n \cdot x)$ converges. This is just sequential compactness of $[0,1]$. But as there are as many $x \in [0,1]$ as there are increasing integer sequences, conceivably the reverse could also hold, though it seems unlikely: and (added) this answer shows it is in fact false.
A: Let $\{a_n\}$ be an strictly increasing sequence of integers. Then $\{\sin(a_n\,x)\}$ does not converge almost everywhere. Suppose to the contrary that $\{\sin(a_n\,x)\}$ converges almost everywhere. Then $(\sin(a_{n+1}x)-\sin(a_nx))^2$ converges almost everywhere to $0$. By the dominated convergence theorem
$$
\lim_{n\to\infty}\int_0^{2\pi}(\sin(a_{n+1}x)-\sin(a_nx))^2\,dx=0,
$$
a contradiction since
$$
\int_0^{2\pi}(\sin(a_{n+1}x)-\sin(a_nx))^2\,dx=2\,\pi.
$$
