Hello, gwammy!

This is a tricky problem, requiring special treatment.

Solve for $\displaystyle x\!:\;\;k \:=\:\frac{\sin x}{1- \cos x}$ We have: .$\displaystyle k - k\cos x \:=\:\sin x \quad\Rightarrow\quad k\cos x + \sin x \:=\:k$

Divide by $\displaystyle \sqrt{1+k^2}\!:\quad \cos x \cdot\frac{k}{\sqrt{1+k^2}} \;+\; \sin x\cdot\frac{1}{\sqrt{1+k^2}} \;\;=\;\;\frac{k}{\sqrt{1+k^2}} $ .[1]

Consider angle $\displaystyle \theta$ in a right triangle with: .$\displaystyle opp = 1,\:adj = k$ Code:

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√1+kČ * |
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* θ |
* - - - - - - - - *
k

The hypotenuse is $\displaystyle \sqrt{1+k^2}$

And: .$\displaystyle \cos\theta = \frac{k}{\sqrt{1+k^2}}, \quad \sin\theta = \frac{1}{\sqrt{1+k^2}}$ .[2]

. . Note that: .$\displaystyle \theta \:=\:\arccos\left(\frac{k}{\sqrt{1+k^2}}\right) $ .[3]

Substitute [2] into [1]: .$\displaystyle \cos x \cos\theta + \sin x\sin\theta \:=\:\frac{k}{\sqrt{1+k^2}}$

. . And we have: .$\displaystyle \cos(x - \theta) \;=\;\frac{k}{\sqrt{1+k^2}} $

. . Then: .$\displaystyle x-\theta \:=\:\arccos\left(\frac{k}{\sqrt{1+k^2}}\right) \quad\Rightarrow\quad x \;=\;\arccos\left(\frac{k}{\sqrt{1+k^2}}\right) + \theta $

Substitute [3]: .$\displaystyle x \;=\;\arccos\left(\frac{k}{\sqrt{1+k^2}}\right) + \arccos\left(\frac{k}{\sqrt{1+k^2}}\right) $

. . Therefore: . $\displaystyle x \;=\;2\arccos\left(\frac{k}{\sqrt{1+k^2}}\right)$