What is the value of θ, if x = 2(θ − sen θ)

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Homework Help Overview

The discussion revolves around converting parametric equations to Cartesian form, specifically focusing on the equations x = 2(θ − sen θ) and y = 2(1 − cos θ). Participants are exploring the relationship between the variables and the parameter θ.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to isolate θ from the equation x = 2(θ − sen θ) and are questioning the meaning of "sen θ," with some clarifying that it refers to the sine function. There is also a suggestion to express the trigonometric part in terms of the variables and to use trigonometric identities.

Discussion Status

The discussion is ongoing, with participants providing clarifications and suggestions for approaching the problem. There is no explicit consensus on a method, but some guidance has been offered regarding expressing variables and identities.

Contextual Notes

There is a mention of potential language differences in the terminology used for the sine function, which may affect understanding. Additionally, participants are grappling with the challenge of solving for θ in terms of y before substituting back into the equation for x.

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Homework Statement



How may I convert this from parametric to cartesian equation?

Homework Equations



x = 2(θ − sen θ)
y = 2(1 − cos θ)

The Attempt at a Solution



I have a problem finding θ from x = 2(θ − sen θ), I don't know how to do it..
 
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sen θ?

Express the trigonometric part in terms of the variables. Then use trigonometric identity.
 
What is sen θ?
Do you mean sin(θ)?
 
Seno = sine in Spanish, as well as Italian and Piedmontese according to Wikipedia.
 
dako said:

Homework Statement



How may I convert this from parametric to cartesian equation?

Homework Equations



x = 2(θ − sen θ)
y = 2(1 − cos θ)

The Attempt at a Solution



I have a problem finding θ from x = 2(θ − sen θ), I don't know how to do it..
Solve for θ in terms of y, then plug that into the equation for x. It's still not pretty.
 

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