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After having shown that

$$f(x) = \frac{x - \sin (x)}{1 - \cos (x)}$$

is strictly monotonically increasing on $x \in (0, 2\pi)$, I'd like to show the surjectivity of this function [to $(0,\infty)$]. However, I fail solving the equation

$$y =\frac{x - \sin (x)}{1 - \cos (x)}$$

with respect to $x$. Can I somehow use/apply the fact that $f(x) = \frac{u(x)}{u'(x)}$ ? I recognized this, even though I could not use it proving the monotonicity, so maybe I can use it here. For any hints I am very thankful.

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    $\begingroup$ the domain is $(0, 2\pi)$, so it is monotone increasing $\endgroup$ – Harambe Nov 5 '17 at 10:11
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Just show that the function is continuous in the domain, and since it tends to $0$ as $x \rightarrow 0$, and tends to $\infty$ as $x \rightarrow 2\pi$, you are done as you have already shown it's monotonically increasing in the domain.

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    $\begingroup$ Instead of "equals $0$ at $x=0$", it should rather be "tends to $0$" as $x\to 0$". $\endgroup$ – Hagen von Eitzen Nov 5 '17 at 11:23
  • $\begingroup$ @HagenvonEitzen Thanks for the correction, I edited it. $\endgroup$ – ab123 Nov 5 '17 at 11:39

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