If $\sin c_n= 3c_n-a_n$ and $a_n\to0$, then $ c_n \to0$

Question

Let $(a_n)_n, (b_n)_n\subseteq \mathbb{R}^{*}$ such that, for every $n$, $$\sin(a_n+b_n)=2a_n+3b_n$$ Show that, if $ a_n \xrightarrow[n \to \infty]{} 0$ then $ b_n \xrightarrow[n \to \infty]{} 0$.

I denote $a_n+b_n=c_n$. Then $c_n$ is the unique solution of the equation $\sin \ x= 3x-a_n$.

I want to prove that $ c_n \xrightarrow[n \to \infty]{} 0$, but I don't know how to continue.

Answer 1

Note that $$ \sin(a_n+b_n)=2a_n+3b_n $$ implies $$ 3|b_n| = |\sin(a_n+b_n)-2a_n|\le |\sin(a_n+b_n)| + 2|a_n| \le 3|a_n| + |b_n| $$ since $|\sin x|\le|x|$ for every $x \in \Bbb R $. Therefore, $$ 2|b_n| \le 3|a_n| $$ so that, if $ a_n \to 0$, then $b_n \to 0$.

Answer 2

First note that $c_n$ is bounded. If $c_{n_k}\to\ell $ on a subsequence $n_k $, then $a_{n_k}\to 0$ hence $\sin\ell=3\ell$, thus $\ell=0$, because $0$ is the only solution of $\sin(x)=3x$. This proves, in particular, $\limsup c_n=\liminf c_n=0$ hence $c_n\xrightarrow{n\to\infty}0$.

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Note that the argument above can by applied directly to the sequence $b_n$.