Find all $a>0$, so that the equation $x=a\sin x+b$ has exactly one solution in $[0,a+b]$, ($b>0$). For what $a>0$ is there exactly one solution in $[0,a+b]$ to the equation $$x=a\sin x+b$$ for all $b>0$.
Thank you,
Eitan. 
 A: Let $f(x)=a\sin x+b-x$.

Here I show that if $a>1$, then there exists a $b>0$ for which the equation $f(x)=0$ has multiple solutions on $[0,a+b]$.
Let $\bar{x}(\varepsilon)=2\pi-\varepsilon$, where $\varepsilon\in(0,\pi)$. We have $$f(\pi)=b-\pi>0,\qquad f(\bar{x}(\varepsilon))=b-a\sin(\varepsilon)+\varepsilon-2\pi\leq0$$ and $$f(2\pi)=b-2\pi\geq 0 $$
if $$2\pi\leq b\leq a\sin(\varepsilon)-\varepsilon+2\pi.$$ As long as $$a\geq \frac{\varepsilon}{\sin\varepsilon}=\bar{a}(\varepsilon),$$ there exists such a $b$, and by the intermediate value theorem there must be at least two solutions to $f(x)=0$ in $(\pi,2\pi]\subseteq[0,a+b]$. 
Finally, note that $\bar{a}(\varepsilon)>1$ for all $\varepsilon\in (0,\pi)$ while $\lim_{\varepsilon\to0^+}\bar{a}(\varepsilon)=1$.

Now I show that if $a\in(0,1]$ then the solution is unique for all $b>0$.
For all $x$ we have $$f'(x)=a\cos x-1\leq 0$$
with strict inequality for all $x$ if $a<1$ and strict inequality when $a=1$ at all but isolated points ($x=0,2\pi,\ldots$). Thus $f$ is strictly decreasing on $[0,\infty)$. 
For all $b>0$ we have $$f(0)=b>0$$ and $$f(a+b)=a\sin(a+b)-a\leq 0. $$ 
By the intermediate value theorem there exists at least one solution to $f(x)=0$ in $[0,a+b]$ and it is unique because $f$ is strictly decreasing.
