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

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The equation R(tanx + 1/cosx + x - 1) = 1/2R2(tanx - x) has been analyzed, revealing that x=0 appears to be a solution across all values of R, though counterexamples exist, such as R=0.3, where a root near x=0.6 is found. Rearranging the equation indicates that for any x in the interval (0, pi/2), a corresponding R can yield a solution. Additionally, a separate inquiry regarding the parametric form of a point wrapping around a unit circle from t=0 to t=a remains unresolved, with participants expressing uncertainty about its nature, speculating it might be a logarithmic spiral. The discussion highlights the complexity of solving the original equation and the challenges in understanding the parametric path. Overall, the conversation reflects a blend of analytical and graphical approaches to mathematical problem-solving.
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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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