Ytukyg

I have already prove that it is true if exist $sin(x_m) = 1$ or $sin(x_m) = -1$. So I need make proof only for absolute-decreasing progression. But I am afraid that $sin(x_n)$ can has limit $1$ and can go to $1$ very fast. Is it a good way to count limit $sin(x_n)$? And what solution is correct?

Let $y_n = |x_n|$

Then $y_{n+1} = y_n |sin y_n| leq y_n forall n$ (*)

So $y_n$ is decreasing and bounded below by $0$. Therefore it converges to some limit $l$.
By passing to the limit in (*) we get that either $l=0$ or $sin l= 1$ giving the family of solutions discussed in the comments, namely $l=2kpi+frac{pi}{2}$

Now note that if $y_n$ converges to $0$ then $x_n$ also converges to $0$. If $y_n$ converges to a nonzero value $l$ then a little more work will also prove that $x_n$ also converges to either $l$ or $-l$ because for n large enough $y_n$ will be between $l$ and $l+frac{pi}{2}$ hence $x_n$ will have constant sign

Either way, $x_n$ will also be convergent

Note that $0le|x_{n+1}|=|x_nsin x_n|le|x_n|$, so the sequence of absolute values, $|x_0|,|x_1|,|x_2|,ldots$ definitely converges to a nonnegative limit $L(x_0)$ satisfying the equation $L(x_0)=L(x_0)|sin L(x_0)|$. If $L(x_0)=0$, we're done: $|x_n|to0$ implies $x_nto0$. If $L(x_0)gt0$, we need only worry about the possibility that $x_n$ approaches both $L(x_0)$ and $-L(x_0)$. But this can't happen: If $x_napproxsigma L(x_0)$ where $sigma=sin L(x_0)in{1,-1}$, then, using the fact that $sin(sigma u)=sigmasin u$ for all $u$ and $sigma^2=1$, we have

$$x_{n+1}=x_nsin x_napproxsigma L(x_0)sin(sigma L(x_0))=sigma L(x_0)(sigmasin L(x_0))=sigma L(x_0)(sigma^2)=sigma L(x_0)approx x_n$$

Remarks: As ImNotTheGuy observes in a comment beneath the OP, $pi/2+2kpi$ (with $kinmathbb{Z}$) is a fixed point for the mapping $f(x)=xsin x$, which means there are sequences that tend to each such limit, as well as sequences that tend to $0$. There are, in fact, uncountably many sequences tending to each of the possible limits: If $kge0$, for example, and $x_n=pi/2+2kpi+epsilon$ with $epsilongt0$ but very small, then $x_{n+1}=x_ncosepsilon =pi/2+2kpi+epsilon'$ with

$$epsilon'=epsiloncosepsilon-(1-cosepsilon)(pi/2+2kpi)approxepsilon-(epsilon^2/2)(pi/2+2kpi)ltepsilon$$

(but still positive), so each positive limit has a "basin of attraction" that includes an open inteval of values for $x_0$ slightly greater than the limit (and likewise, for each negative limit, an open interval of values slightly less than the limit -- since $|x_{n+1}|=|x_nsin x_n|le|x_n|$, successive terms in any sequence always get closer to the origin).

In general If $x_0$ is an initial approximation of a fixed point , the fixed point iteration generates a sequence of approximants by $$x_{n+1}=varphi (x_n),$$ where $varphi$ is a continuous function. Suppose that at the fixed point that we indicate with $alpha$ we have $$varphi '(alpha)=varphi ''(alpha)=cdotsvarphi ^{p-1}(alpha)=0, quad varphi ^p(alpha)neq 0 qquad (*);$$ then we have the following theorem:

Let $alpha$ be a fixed point of $varphi$ and $I_{epsilon}:={xinBbb R: |x-alpha|<epsilon}$. Asuume that $varphi in Bbb C^p[I_{epsilon}]$ satisfes $(*)$. If $$M(epsilon):=|varphi '(t)|<1$$ then the fixed point iteration converges to $alpha$ for every $x_0in I_{epsilon}$ and the order the convergence is $p$.

Note that to use this theorem you have to know tha fxed point $alpha$ (you can plot the function for example and obtain an approximation of point). You can find this argument in this book.