Solving Equation with Bisection Method - Real Root Between -2*pi and 2*pi

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SUMMARY

The discussion focuses on solving a specific equation using the bisection method, targeting the real root located between -2π and 2π. The equation presented is unbalanced in terms of parentheses, which raises concerns about its validity. The user aims to find one real root while disregarding the two imaginary roots. The bisection method is chosen due to the known interval containing the real root.

PREREQUISITES
  • Understanding of the bisection method for root-finding
  • Familiarity with trigonometric functions and their properties
  • Basic knowledge of complex numbers and roots
  • Ability to manipulate and analyze algebraic equations
NEXT STEPS
  • Study the implementation of the bisection method in Python using libraries like NumPy
  • Learn about the significance of balanced equations in mathematical modeling
  • Explore the properties of trigonometric functions in root-finding scenarios
  • Investigate methods for validating equations before applying numerical techniques
USEFUL FOR

Mathematicians, engineering students, and software developers interested in numerical methods for solving equations and analyzing mathematical models.

jrm2002
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I need to solve this equaation:

(x1cos(p)+y1sin(p))^2 * (x2sin(p)-y2cos(p)) = (x1sin(p)-y1cos(p)) * (x2cos(p)+y2sin(p))^2 ;

x1,x2,y1,y2 are constants

The equation would have 3 roots

1 real and two imaginary

I don't need imaginary roots.

I am planning to use bisection method-----

since i know the real root will lie between -2*pi and 2*pi

What is your comment on this?
 
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Without commenting on your method I see that the equation as stated doesn't make sense. You have unbalanced parentheses on both sides.
 

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