Trig inequalities

1. The problem statement, all variables and given/known data
find all values of x in the interval [0, 2pi] that satisfy the equation 2cosx + 1 less than or equal to 0


2. Relevant equations
None.

3. 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?
 
125
0
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?
 
125
0
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?
 
125
0
Oh, well, technically speaking, if your inequality was [itex] \leq [/itex], then you should include the endpoints in the interval. If your inequality was [itex] < [/itex], 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?
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top