Solve Trig Inequalities: Find x in [0,2pi] for 2cosx+1≤0

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Homework Help Overview

The problem involves solving the inequality 2cos(x) + 1 ≤ 0 for values of x within the interval [0, 2π]. Participants are exploring the conditions under which the cosine function yields values that satisfy this inequality.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find values of x by first solving the equation 2cos(x) + 1 = 0, identifying specific roots. They then question how to determine the intervals where the inequality holds true.
  • Some participants suggest testing values within the intervals defined by the roots to ascertain where the inequality is satisfied.
  • Further discussion includes clarifying the implications of using ≤ versus < in the context of interval inclusion.

Discussion Status

The discussion is active, with participants exploring different intervals and confirming findings. There is a focus on understanding how to approach inequalities and the significance of endpoints in the intervals based on the type of inequality.

Contextual Notes

Participants are working under the constraints of the specified interval [0, 2π] and are considering the implications of including or excluding endpoints based on the inequality type.

EL ALEM
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Homework Statement


find all values of x in the interval [0, 2pi] that satisfy the equation 2cosx + 1 less than or equal to 0


Homework Equations


None.

The Attempt at a Solution


for when 2cosx+1=0
cosx=-1/2
x= 2pi/3, 4pi/3

but what about the values for when 2cosx + 1 is less then 0? How to i fins those?
 
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Since you have the roots - why don't you pick values between [0, 2pi/3), (2pi/3, 4pi/3), and (4pi/3, 2pi] to check? One or more of these intervals will give you your desired answer.
 
Ok so i found 2cosx + 1 < 0 at this interval (2pi/3, 4pi/3) , so would that be my answer?
 
If that is the only interval, then indeed it is! In these types of problems, you can always divide the set into intervals and find test points...
 
Thanks a bunch, one more question, if it was only less than instead of less than or equal to how would we divide the set into intervals?
 
Oh, well, technically speaking, if your inequality was \leq, then you should include the endpoints in the interval. If your inequality was &lt;, then you should exclude the endpoints. So what you're really working with is [2pi/3, 4pi/3].
 
Ok thanks.
 
Ok so i found 2cosx + 1 < 0 at this interval (2pi/3, 4pi/3) , so would that be my answer?
 

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