Discover the Single Solution for 3sinx - 1 = b in (0,2pi): Find the Value of b!

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Homework Help Overview

The discussion revolves around the equation 3sinx - 1 = b, where b is a positive real number, and the goal is to determine the conditions under which this equation has a single solution in the interval (0, 2π).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore how to determine the number of solutions by analyzing the graph of the function 3sinx - 1 and question which parts of the graph indicate the number of solutions. There are discussions about the implications of the interval being open versus closed and how to graph equations that equal a constant.

Discussion Status

Participants are actively engaging with the problem, sharing hints and exploring graphical interpretations. There is a recognition of the need to focus on the intersections of the graph with horizontal lines representing different values of b. Some participants express confusion about the graphing process, while others provide clarifications.

Contextual Notes

There is an emphasis on the specific interval (0, 2π) for the solutions, and participants are navigating the implications of this constraint in their reasoning.

TyErd
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the equation 3sinx - 1 = b, where b is a positive real number, has one solution in the interval (0,2pi). The value of b is:

Frankly I have no idea where to even start with this problem.
 
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Hi TyErd! :smile:

(have a pi: π :wink:)

Hint: how many solutions in (0,2π) does 3sinx - 1 = 15 have?

And 3sinx - 1 = 0.5 ? :wink:
 
Um. thnx for the pi. How do you know how many solutions there are?
 
TyErd said:
Um. thnx for the pi. How do you know how many solutions there are?

Draw a graph and find out.

Get on with it!
 
alright I've drawn the graph, so what parts of the graph do i have to look at to know how many solutions there?
 
hm..
 
Which parts? Umm...

If I asked you to find out how many solutions there are to [itex]x^2=10[/itex] how would you go about doing that by looking at a graph?
 
TyErd said:
alright I've drawn the graph, so what parts of the graph do i have to look at to know how many solutions there?
And you should be looking only at the part of the graph on the interval [0, 2π].
 
Nooo … (0,2π). :wink:
 
  • #10
tiny-tim said:
Nooo … (0,2π). :wink:
Right, tiny-tim. I missed that it was the open interval.
 
  • #11
so when it says solutions, should i be looking at the x intercepts?
 
  • #12
TyErd said:
alright I've drawn the graph, so what parts of the graph do i have to look at to know how many solutions there?

What is the graph that you have drawn? Specifically, what is the formula of the function you have graphed?
 
  • #13
I tried to draw 3sinx-1=15 by taking 15 onto the other side but I've just realized that isn't right. How do you graph an equation that equals a number?
 
  • #14
Graph [itex]15=3sinx-1[/itex]
Then graph [itex]y=15[/itex]

Do you see how if you tried to solve these two equations simultaneously, you would get [itex]15=3sinx-1[/itex]?

Are there any intersections between the two functions?

Now, can you find any number b (such as 15) that makes it such that the equation [itex]3sinx-1=b[/itex] has only 1 real solution in the interval between 0 and 2π?
This is the same as saying for what number b will the function [itex]y=b[/itex] intersect the function [itex]y=3sinx-1[/itex] only once in the interval [itex](0,2\pi)[/itex]?
 
  • #15
Thankyouuu! i get it finally, you made it so much easier. thnx
 

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