Proving a Single Solution for an Equation between 0 and pi/2

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Homework Help Overview

The discussion revolves around finding solutions to a specific equation involving trigonometric functions within the interval from 0 to pi/2. The equation presented is R(tanx + 1/cosx + x - 1) = 1/2R2(tanx - x), and participants are exploring the nature of its solutions and the possibility of proving uniqueness.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the potential for algebraic methods to solve the equation and question the presence of multiple solutions based on different values of R. There is also a mention of using graphing calculators for insights into the solutions.

Discussion Status

The conversation is ongoing, with some participants expressing skepticism about the original poster's claim of x=0 being the only solution. Alternative viewpoints regarding the existence of other solutions for different values of R have been introduced, indicating a productive exploration of the problem.

Contextual Notes

There are indications of confusion regarding variable usage in the equation, as well as the nature of the functions involved. Participants are also considering the implications of rearranging the equation to analyze solutions further.

turdferguson
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The last step in a problem of mine is to find all solutions between 0 and pi/2 of the following equation:
R(tanx + 1/cosx + x - 1) = 1/2R2(tanx - x)

By simple inspection and with the aid of a graphing calculator, its obvious that the only solution is x=0, regardless of the value of R. This is enough for the assignment, but I am wondering if there's a way to mathematically isolate x and prove that 0 is the only solution
 
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With t both inside and outside the trig functions (and do you really mean to have t on one side and x on the other?), there is no "algebraic" method of solving this equation.
 
Sorry about the t...

Thanks, I thought there might have been a useful identity, but it looks like the graphing calculator will do.
 
turdferguson said:
The last step in a problem of mine is to find all solutions between 0 and pi/2 of the following equation:
R(tanx + 1/cosx + x - 1) = 1/2R2(tanx - x)

By simple inspection and with the aid of a graphing calculator, its obvious that the only solution is x=0, regardless of the value of R.

Not true. For example when R = 0.3 there's a root close to x = 0.6

If you rearrange it as (tan x - x ) / (tan x + 1/cos x + x - 1) = 2R^3, then it's obvious that for any value of x=X between 0 and pi/2 there is some value of R that makes X a solution of the equation.
 
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Ive got another question, so Ill put it in here. As we know, the wrapping function takes a point (1,a) on a number line and wraps it about a unit circle until it gets to the point (cosa, sina).

Today on a test, we were asked to find the parametric form of the path of the point from t=0 to t=a. No one got it and well go over it next week, but I am anxious to know how to solve something like this. I've only dealt with parametric lines and the path is obviously not a line. I started to mess around with polar graphs in my calculator, but nothing comes close. Is it a logarithmic spiral?
 
Ive got another question, so Ill put it in here. As we know, the wrapping function takes a point (1,a) on a number line and wraps it about a unit circle until it gets to the point (cosa, sina).

Today on a test, we were asked to find the parametric form of the path of the point from t=0 to t=a. No one got it and well go over it next week, but I am anxious to know how to solve something like this. I've only dealt with parametric lines and the path is obviously not a line. I started to mess around with polar graphs in my calculator, but nothing comes close. Is it a logarithmic spiral?
 

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