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I have an equation to be solved. But Mathematica does not work for it. I hope the solution x can be expressed as a function of a and b

Solve[(x/Sin[a])^2 == ((1 - x)/Sin[b])^2 + 1 + (x/Tan[a] - (1 - x)/Sin[b] Cos[b])^2 - 2 ((1 - x)/Sin[b]) Sqrt[1 + (x/Tan[a] - (1 - x)/Sin[b] Cos[b])^2] Sin[b - ArcTan[x/Tan[a] - (1 - x)/Sin[b] Cos[b]]], x]

By using $x = \frac{y}{l}$, the above equation is reduced from the following one:

Solve[(y/Sin[a])^2 == ((l - y)/Sin[b])^2 + l^2 + (y/Tan[a] - (l - y)/Sin[b] Cos[b])^2 - 2 ((l - y)/Sin[b]) Sqrt[l^2 + (y/Tan[a] - (l - y)/Sin[b] Cos[b])^2] Sin[b - ArcTan[(y/Tan[a] - (l - y)/Sin[b] Cos[b])/l]], y]

If the reduced equation can be simplified to zero, how about this original one? I cannot simplify the original one to zero. From this original equation, can we get y as a function of l, a and b?

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  • $\begingroup$ Mathematica solves your equation and gives {}: no solution! $\endgroup$ – Ulrich Neumann Jul 1 at 20:47
  • $\begingroup$ Yes, why is there no solution? In principle, there should be a solution x as a function of a and b. I do not know why Mathematica gives { }. $\endgroup$ – Hao Wu Jul 1 at 20:50
  • $\begingroup$ Solve probably fails to get a "proper" simplification. It is actually tautologically true (that is, for all values of x). $\endgroup$ – Daniel Lichtblau Jul 1 at 20:56
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Your equation does not contain any information:

(x/Sin[a])^2 == ((1 - x)/Sin[b])^2 + 1 + (x/Tan[a] - (1 - x)/Sin[b] Cos[b])^2 -
  2 ((1 - x)/Sin[b]) Sqrt[1 + (x/Tan[a] -
  (1 - x)/Sin[b] Cos[b])^2] Sin[b - ArcTan[x/Tan[a] -
  (1 - x)/Sin[b] Cos[b]]] // FullSimplify

True

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  • $\begingroup$ Thank you! How about the original one? In terms of your answer, I modify my post. $\endgroup$ – Hao Wu Jul 1 at 20:59
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    $\begingroup$ If you replace Solve with FullSimplify in the original one it also gives True, though you need to add the assumption that l > 0. $\endgroup$ – bill s Jul 2 at 1:25

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