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I'm looking for a solution to $b\sin({\theta})\cos({\phi})+a\cos({\theta})\sin({\phi})=0$ for the variable $\phi$.

In the equation both $a$ and $b$ are real numbers; in particular, I have $\frac{a}{b}>1$.

I'm looking for a solution for $\phi$ in the interval $\phi \in [0,\pi]$.

$\theta$ is also a real parameter which varies in the interval $\theta \in [0,\pi]$ (the same of $\phi$).

I'm trying to solve this equation with mathematica with the command "Reduce", I attach a part of the code:

 k[\[Theta]_, \[Phi]_] := -b Cos[\[Phi]] Sin[\[Theta]] - 
  a Cos[\[Theta]] Sin[\[Phi]]

Assuming[a \[Element] Reals && a > 0 && b \[Element] Reals && b > 0, 
Reduce[k[\[Theta], \[Phi]] == 0, \[Phi], 
 Reals]]

But of course I get all the periodic solutions as results and they're hard to interpret. Is there a way to restrict the solutions only in the intervals I'm interested in? Or is there a quick solution to this problem?

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1 Answer 1

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We have $a\sin\theta\cos\phi=-b\cos\theta\sin\phi$

$\iff\dfrac{\sin\phi}{\cos\phi}=-\dfrac ab\dfrac{\sin\theta}{\cos\theta}\iff \tan\phi= -\dfrac ab\tan\theta$

$\phi=n\pi+\arctan\left(-\dfrac ab\tan\theta\right)$ where $n$ is any integer

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