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With this proof, I need to prove two things.

1) I need to prove the equation has solutions

2) Show the solution is unique

I know to prove the equation has solutions, I would like to use the intermediate value theorem, but I don't know exactly how to incorporate it. For number two, I figured how to prove the solution is unique , by some help. I used the Mean Value Theorem to prove it, but I would like to know how to prove it using the Rolle's Theorem. I would like to see it double sided, but I have never really went into depth on the Rolle's Theorem.

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  • $\begingroup$ When your question is put on hold, don't post a new version, edit the already existing one. $\endgroup$ – Daniel Fischer Dec 16 '16 at 12:04
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Let $y=x-a\sin x -b$, then the above equation has a uniques solution iff $y$ has a unique root.

Note $y'=1-a\cos x> 0$, hence $y$ is an increasing function, this already shows that if a solution exists, it has to be unique.

Now we use Intermediate Value Theorem to prove the existence of a root. For $x>a+b+1$, we have $y>a+b+1-a-b=1$. For $x<-a+b-1$, we have $y<-a+b-1+a-b=-1$, hence IVT implies that we have a root of $y$ in $(-a+b-1,a+b+1).$

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It is equivalent to show that $f(x)=x-a\sin x$ is a bijection of $\mathbb{R}$. Since $x\mapsto x$ is a bijection and $x\mapsto a\sin x$ is a bounded continuous function, $f$ is clearly surjective. To show $f$ is also injective just notice that $$ f'(x)=1-a\cos x\geq1-a>0,\,\,x\in\mathbb{R}. $$

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Let $x_1$ and $x_2$ be solutions of the above equation and assume that $x_1 \ne x_2$.

Then we have

$x_1-x_2=a(\sin x_1- \sin x_2)$

or

$1=a \frac{\sin x_1- \sin x_2}{x_1-x_2}$. Therefore $x_1=x_2$.

By the Mean Value Th., there is $t $ between $x_1$ and $x_2$ such that

$1=a \cos t$. Hence $1/a= \cos t$. But this is impossible since $1/a>1$.

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