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?

Consider the function:
$$
f(x)=x-a\sin(x)-b
$$
This has derivative:
$$
f'(x)=1-a\cos(x)>0
$$
since $0<a<1$, so $f(x)$ is increasing on $\mathbb{R}$. Hence as $f(x)$ is continuous increasing and is negative for sufficiently negative $x$ and positive for sufficiently positive $x$ it has a zero on $\mathbb{R}$, and it is unique.
