Prove that $x-a \sin(x)=b$ has one real solution, where $0\lt a \lt 1 $ $a,b \in \mathbb{R}$. Prove that $x-a\sin(x)=b$ has one real solution, where $0\lt a \lt 1 $.
I need some sort of starting hint as to how to prove this.
I can define $g(x)= x-a\sin(x)-b$ but more than that I'm having difficulties proving. What theorem could I use?

Its derivative is always positive, strictly, since $$|asin(x)| \leq |a| <1.$$
So it has at most one zero.
When $x$ goes to infinities, we have positive and negative limits, meaning, by intermediate value theorem, that it does have one zero.
