Finding a Solution for a System of Equations with Trigonometric Functions

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SUMMARY

The discussion focuses on solving a system of equations involving trigonometric functions, specifically the equations y = k²(1 - cos(θ))/2 and x = k²(θ - sin(θ))/2. The user seeks to find a value of k that satisfies the equations for a given point (x₀, y₀). After manipulating the equations, they derived the expression 2(θ - sin(θ)) = 1 - cos(θ) but encountered difficulties in determining the value of θ. The suggested method for finding θ involves iterative trial solutions, emphasizing the importance of using radians.

PREREQUISITES
  • Understanding of trigonometric functions and their properties
  • Familiarity with algebraic manipulation of equations
  • Knowledge of iterative methods for solving equations
  • Basic understanding of radians and their application in trigonometry
NEXT STEPS
  • Research iterative methods for solving nonlinear equations
  • Learn about numerical methods for finding roots, such as the Newton-Raphson method
  • Explore the properties of trigonometric functions and their graphs
  • Study the application of radians in trigonometric calculations
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Students studying mathematics, particularly those focusing on trigonometry and algebra, as well as educators looking for methods to teach solving systems of equations with trigonometric functions.

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Homework Statement


It's a longer problem but all that's left is:

y = k^2 \frac{1-cos(\theta)}{2}
x = k^2 \frac{\theta - sin(\theta)}{2}

I want to find a k that solves the equations for the point (x, y) = (x_0, y_0)

The Attempt at a Solution


I manipulated them to get two expressions for k, putting them = each other and simplifying I get

2(\theta-sin(\theta)) = 1 - cos(\theta)
And now I'm completely stuck, how on Earth do I go about finding what theta is?
 
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You can always iterate, picking a trial solution value for theta: remember to use radians.
 
Yeah, that's an option of course. I wonder if that's what the book did, would mean that theta is some relatively nice number I could just find by brute forcing it.
Will give it a go tomorrow, calling it a day for now.
 

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