Prove without using graphing calculators that $f: \mathbb R\to \mathbb R,\,f(x)=x+\sin x$ is both one-to-one, onto (bijective) function. Prove that the function $f:\mathbb R\to \mathbb R$ defined by $f(x)=x+\sin x$ for $x\in \mathbb R$ is a bijective function.

The codomain of the $f(x)=x+\sin x$ is $\mathbb R$ and the range is also $\mathbb R$. So this function is an onto function.
But I am confused in proving this function is one-to-one.
I know about its graph and I know that if a function passes the  horizontal line test (i.e horizontal lines should not cut the function at more than one point), then it is a one-to-one function. The graph of this function looks like the graph of $y=x$ with sinusoids going along the $y=x$ line.
If I use  a graphing calculator at hand, then I can tell that it is a one-to-one function and $f(x)=\frac{x}{2}+\sin x$ or $\frac{x}{3}+\sin x$ functions are not, but in the examination I need to prove this function is one-to-one theoritically, without graphing calculators.
I tried the method which we generally use to prove a function is one-to-one but no success.
Let $f(x_1)=f(x_2)$ and we have to prove that $x_1=x_2$ in order fot the function  to be one-to-one.
Let $x_1+\sin x_1=x_2+\sin x_2$
But I am stuck here and could not proceed further.
 A: Assume $f$ is Many-One.
So there do exist some $x_{1}$ and $x_{2}$ where $x_{1}\neq x_{2}$ such that $$f(x_{1})=f(x_{2})$$
$$x_{1}+\sin x_{1}=x_{2}+\sin x_{2}$$$$x_{1}-x_{2}=\sin x_{2}-\sin x_{1}=-2\cos\left(\frac{x_{1}+x_{2}}{2}\right)\sin\left(\frac{x_{1}-x_{2}}{2}\right)$$
$$\left|x_{1}-x_{2}\right|=2\left|\cos\left(\frac{x_{1}+x_{2}}{2}\right)\sin\left(\frac{x_{1}-x_{2}}{2}\right)\right|\leq2\left|\sin\left(\frac{x_{1}-x_{2}}{2}\right)\right|$$
$$\implies\left|\frac{\sin\left(\frac{x_{1}-x_{2}}{2}\right)}{\frac{x_{1}-x_{2}}{2}}\right|\geq 1$$
which is obviously a contradiction.
So our assumption that $f(x)$ is Many-One is false.
$f(x)$ being onto is trivial since on the RHS $x$ can take all real values. 
A: You can prove that this function is strictly increasing :
It's a $C^1$ function and $f'(x) = 1+\cos(x) \geq 0$, so the function is increasing.
$\{x \mid f'(x) = 0 \} = \pi \Bbb Z$ is a discrete set, so $f$ is strictly increasing (if $f$ was locally constant somewhere, there would be an intervall $]a,b[$ where $f'(x)=0$ )
