I am trying to solve the equation $$\tan ( \frac{\alpha}{2} ) - \frac{d + r - r \cos(\alpha)}{z + r \sin(\alpha) + \alpha r} = 0$$ all morning. Actually this is the result from a much more complex equation. But it seems to be correct (I tried it with online tools).
I know a few things about this equation: $$r > 0 \qquad d > 0 \qquad z > 0$$ Only the interval $$0 < \alpha < \pi$$ is interesting for me. I am pretty sure that there is exact one root point.
Do you think it is possible to proof, there is exactly one root point in this interval? Calculating the root would be nice, but not necessary, because it is sufficient for me to proof there is exactly one root point to continue with my main proof. Do you have any ideas, how I can continue? Thank you very much!
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EDIT
If we look at $$tan(\frac{\alpha}{2}) - A(\alpha) = 0$$ It should be enough to know that $A(\alpha)$ is well defined over the interval , has at least one positive value in the interval and that is would not "jump" (that it is continuous). If $A(\alpha)$ has these properties, $tan(\frac{\alpha}{2}) - A(\alpha) = 0$ has exactliy one root point in the interval $0 < \alpha < \pi$. Is this correct?