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While working on an engineering problem, we have met a non linear system of two variables and two equations.

$$ \left\{ \begin{array}{c} A=x \sin y \\ B=-x^2 + x \cos y \end{array} \right. $$ where $A$ and $B$ are constants.

if the system is solved analytically two solutions ($Z1$ and $Z2$) can be found from: $(A^2+B^2)+Z^2+Z(2B -1) =0 $

where $Z=x^2$

but if we solve the system with Newton-Raphson method only one solution is found. How to obtain the second one?

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  • $\begingroup$ Just detect where $y $ is poaitive and negative for two separate intervals . You will get two solutions . This can be done only if we know tge exact values of $A,B $. $\endgroup$ – Archis Welankar Oct 4 '17 at 11:20
  • $\begingroup$ The equation giving two solutions for $Z$ is not equivalent to the original system, but obtained by squaring. Are you sure the second solution is a solution of the original system? $\endgroup$ – Professor Vector Oct 4 '17 at 11:38
  • $\begingroup$ Thank you for your answers; @ArchisWelankar the problem is that the two solutions are positive, we need the positive solution for $y1$ and $y2$ and the exact values of $A$ and $B$ are known. $\endgroup$ – Lahouaria Oct 4 '17 at 14:44
  • $\begingroup$ Thank you for your answers; @ProfessorVector we are juste using $Z$ as a change of variable to help find the solutions of the original system. the equation using $Z$ has in fact 4 solutions (2 positive solutions and 2 negative solutions) we only need the two positive ones of the system. yes the second solution is a solution of the original system. $\endgroup$ – Lahouaria Oct 4 '17 at 14:44
  • $\begingroup$ Then you need to determine the nature of the graph (like increasing) or otherwise you need to use to different starting points. $\endgroup$ – Archis Welankar Oct 4 '17 at 15:56

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