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I am interested in finding all the solution $(x_1,x_2,x_3,x_4)\in \mathbb{R}^4$ of equations:

$$\sin \pi x_1\sin\pi x_2=\sin \pi x_3\sin\pi x_4.$$

I have found out a paper: Rational products of sines of rational angles --- GERALD MYERSON, which answers my question partially.

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Here is my interpretation:

Suppose $\alpha,\beta,\gamma,\delta\in \left(0,\dfrac{\pi}{2}\right]\cap \mathbb{Q}\pi$ and $\sin \alpha\sin \beta=\sin\gamma\sin\delta$.

Then $(\alpha,\beta,\gamma,\delta)=$ $(x,y,x,y)$ or $(x,y,y,x)$

or $\left(\dfrac{\pi}{6},\phi,\dfrac{\phi}{2},\dfrac{\pi}{2}-\dfrac{\phi}{2}\right)$ or $\left(\dfrac{\pi}{6},\phi,\dfrac{\pi}{2}-\dfrac{\phi}{2},\dfrac{\phi}{2}\right)$ or $\left(\phi,\dfrac{\pi}{6},\dfrac{\phi}{2},\dfrac{\pi}{2}-\dfrac{\phi}{2}\right)$ or $\left(\phi,\dfrac{\pi}{6},\dfrac{\pi}{2}-\dfrac{\phi}{2},\dfrac{\phi}{2}\right)$

or $\left(\dfrac{\pi}{2}-\dfrac{\phi}{2},\dfrac{\phi}{2},\phi,\dfrac{\pi}{6}\right)$ or $\left(\dfrac{\phi}{2},\dfrac{\pi}{2}-\dfrac{\phi}{2},\phi,\dfrac{\pi}{6}\right)$ or $\left(\dfrac{\pi}{2}-\dfrac{\phi}{2},\dfrac{\phi}{2},\dfrac{\pi}{6},\phi\right)$ or $\left(\dfrac{\phi}{2},\dfrac{\pi}{2}-\dfrac{\phi}{2},\dfrac{\pi}{6},\phi\right)$

or $\left(x_1^0,x_2^0,x_3^0,x_4^0\right)$ or $\left(x_1^0,x_2^0,x_4^0,x_3^0\right)$ or $\left(x_2^0,x_1^0,x_3^0,x_4^0\right)$ or $\left(x_2^0,x_1^0,x_4^0,x_3^0\right)$

or $\left(x_4^0,x_3^0,x_2^0,x_1^0\right)$ or $\left(x_3^0,x_4^0,x_2^0,x_1^0\right)$ or $\left(x_4^0,x_3^0,x_1^0,x_2^0\right)$ or $\left(x_3^0,x_4^0,x_1^0,x_2^0\right)$, where $x,y,\phi\in \left(0,\dfrac{\pi}{2}\right]$ and $(x_1^0,x_2^0,x_3^0,x_4^0)\in \left\{ \left( \dfrac{\pi}{21}, \dfrac{8\pi}{21}, \dfrac{\pi}{14}, \dfrac{3\pi}{14} \right), \left( \dfrac{\pi}{14}, \dfrac{5\pi}{14}, \dfrac{2\pi}{21}, \dfrac{5\pi}{21} \right), \left( \dfrac{4\pi}{21}, \dfrac{10\pi}{21}, \dfrac{3\pi}{14}, \dfrac{5\pi}{14} \right), \left( \dfrac{\pi}{20}, \dfrac{9\pi}{20}, \dfrac{\pi}{15}, \dfrac{4\pi}{15} \right), \left( \dfrac{2\pi}{15}, \dfrac{7\pi}{15}, \dfrac{3\pi}{20}, \dfrac{7\pi}{20} \right), \left( \dfrac{\pi}{30}, \dfrac{3\pi}{10}, \dfrac{\pi}{15}, \dfrac{2\pi}{15} \right), \left( \dfrac{\pi}{15}, \dfrac{7\pi}{15}, \dfrac{\pi}{10}, \dfrac{7\pi}{30} \right), \left( \dfrac{\pi}{10}, \dfrac{13\pi}{30}, \dfrac{2\pi}{15}, \dfrac{4\pi}{15} \right), \left( \dfrac{4\pi}{15}, \dfrac{7\pi}{15}, \dfrac{3\pi}{10}, \dfrac{11\pi}{30} \right), \left( \dfrac{\pi}{30}, \dfrac{11\pi}{30}, \dfrac{\pi}{10}, \dfrac{\pi}{10} \right), \left( \dfrac{7\pi}{30}, \dfrac{13\pi}{30}, \dfrac{3\pi}{10}, \dfrac{3\pi}{10} \right), \left( \dfrac{\pi}{15}, \dfrac{4\pi}{15}, \dfrac{\pi}{10}, \dfrac{\pi}{6} \right), \left( \dfrac{2\pi}{15}, \dfrac{8\pi}{15}, \dfrac{\pi}{6}, \dfrac{3\pi}{10} \right), \left( \dfrac{\pi}{12}, \dfrac{5\pi}{12}, \dfrac{\pi}{10}, \dfrac{3\pi}{10} \right), \left( \dfrac{\pi}{10}, \dfrac{3\pi}{10}, \dfrac{\pi}{6}, \dfrac{\pi}{6} \right) \right\}.$

  1. Is my interpretation correct? Do I miss any possibilities?
  2. What is about the non-rational solutions? Is there any way to solve the equation over $\mathbb{R}$?
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  • $\begingroup$ The question does not make much sense over $\mathbb R$, as for most choices of $x_1,x_2,x_3$, there is a solution $(x_1,x_2,x_3,\pi^{-1}\arcsin(\sin(\pi x_1)\sin(\pi x_2)/\sin(\pi x_3)))$. There is nothing to classify here. $\endgroup$ – Emil Jeřábek Jul 22 '20 at 8:39
  • $\begingroup$ What do you mean by "solving over R"? This equation defines a hypersurface in $R^4$ which evidently contains infinitely many points. What does it mean to solve? Parametrize this hypersurface? $\endgroup$ – Alexandre Eremenko Jul 22 '20 at 12:31
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    $\begingroup$ I'm glad you found my paper. $\endgroup$ – Gerry Myerson Jul 22 '20 at 12:36