Show that, for any constants a ∈ (0, 1) and b ∈ R, the equation x = a sin x + b has a unique solution. I received the first part by using what was stated below and using the Mean Value Thm. And now I have to show that there is a unique solution, I was thinking using the Intermediate Value Thm to prove this??? how do i incorporate it?
 A: Suppose that the equation has two distinct solutions $x,y$ given $a\in(0,1)$. Then, $$|x-y| = a|\sin x - \sin y| = a|\cos\varepsilon||x-y|$$, for some $\varepsilon$ between $x,y$ by the Mean Value Theorem. But this is a contradiction because $a|\cos\varepsilon|<1$. 
This proves that if the equation has a solution, then it is unique. For the existence, note that $f(x)= x-a\sin x$ is an odd function and $\lim_{x\to\infty} = \infty$, so $f(x) = b$ must have a solution for every real $b$.
A: Just intuitively speaking.
For $a$ positive but less than $1$, $y= a \sin{x}$ contact $y=x$ only at $x=0$ and then remains below $y=x$.
When you raise or lower $y= a \sin{x}$ by a constant $b$, then the steepness (derivative) of the line $y=x$ will always be greater than that of $y= a \sin{x}$.
The graph of the sine is confined in a strip of width less than $2$ and is continuous. The line will always cross the sine: but soon after the cross point, because of the lesser derivative, the sine will not succeed and "catch" the line again.
