I have an equation I'd like to solve given as $$s \cdot \alpha \equiv p (\mathrm{mod} 2 \pi) $$ The numbers $s$ and $p$ are known, while $\alpha$ is to be solved for, and additionally is between $0$ and $2 \pi$. I have found a solution using other information, so I know solutions exist, but I'd like to know if solutions are unique and more importantly a general way of calculating them from the given information.
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3$\alpha\equiv \dfrac ps\mod\dfrac{2\pi}{s}$. – Bernard Jul 26 '17 at 22:10
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Solutions are not generally unique. If $|s| > 1$ there can be multiple solutions. Example: $s = 4, p = 0$. Then $\alpha = 0, \pi/2, \pi, 3\pi/2$ are all solutions.
As for a general method: first solve $$ s \alpha = 0 \bmod 2\pi$$ to get a list of alpha values, say $u_1, u_2, \ldots, u_k$.
Then solve $s\alpha = p \bmod 2\pi$ to get ONE alpha-value, call it $v$ (using the method of Bernard's hint, for instance).
Then $$ u_1 + v, u_2 + v, \ldots, u_k + v $$ will all be solutions to your original equation (but you may have to mess around a bit to get them within the range 0 to $2 \pi$.
John Hughes
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$$\begin{align} s\alpha\, &\equiv\, p\!\!\pmod{\!2\pi \Bbb Z}\\[.3em] \iff\ s\alpha\, &=\, p\ +\ 2\pi n,\quad {\rm for\ some}\ \ n\in\Bbb Z\\[.3em] \iff\ \ \ \alpha\, &=\, \dfrac{p}s + \dfrac{2\pi}s\, n,\,\ \ {\rm for\ some}\ \ n\in\Bbb Z\\[.3em] \iff\ \ \ \alpha\, &\equiv\, \dfrac{p}{s}\!\!\pmod{\!\dfrac{2\pi}s\, \Bbb Z} \end{align}$$
Bill Dubuque
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