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

  • Thread starter turdferguson
  • Start date
In summary, the conversation discusses finding all solutions between 0 and pi/2 of the equation R(tanx + 1/cosx + x - 1) = 1/2R2(tanx - x) and whether there is a way to mathematically isolate x and prove that 0 is the only solution. It is concluded that there is no algebraic method to solve the equation and a graphing calculator can be used to find the solution. The conversation also mentions the wrapping function and finding the parametric form of the path of a point on a unit circle.
  • #1
turdferguson
312
0
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
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
Sorry about the t...

Thanks, I thought there might have been a useful identity, but it looks like the graphing calculator will do.
 
  • #4
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.
 
Last edited:
  • #5
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?
 
  • #6
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?
 

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

What is the process for proving a single solution for an equation between 0 and pi/2?

The process for proving a single solution for an equation between 0 and pi/2 is to first solve the equation algebraically. Then, substitute the value of pi/2 into the equation to see if it produces a valid solution. If it does, then the equation has a single solution between 0 and pi/2.

Can there be more than one solution for an equation between 0 and pi/2?

No, an equation between 0 and pi/2 can only have a single solution. This is because the range of values between 0 and pi/2 is limited, and any additional solutions would fall outside of this range.

What is the significance of the range between 0 and pi/2 for proving a single solution?

The range between 0 and pi/2 is significant because it represents the first quadrant of the unit circle in trigonometry. This means that any solutions within this range will correspond to a unique angle measure between 0 and 90 degrees.

Are there any special cases to consider when proving a single solution for an equation between 0 and pi/2?

Yes, there are two special cases to consider. The first is when the equation contains a function with an inverse that has a restricted domain, such as tan inverse. In this case, the solution must be within the restricted domain for it to be valid. The second case is when the equation contains a trigonometric function with multiple periods, such as sine or cosine. In this case, the solution must also fall within the first period (between 0 and pi/2) for it to be considered a single solution.

Can a graphical method be used to prove a single solution for an equation between 0 and pi/2?

Yes, a graphical method can be used to prove a single solution for an equation between 0 and pi/2. This involves graphing the equation and visually identifying the point(s) of intersection with the x-axis within the range of 0 and pi/2. If there is only one point of intersection, then it is a single solution within the given range.

Back
Top