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

[EL ALEM]

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?

 

[cyby]

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.

 

[EL ALEM]

Ok so i found 2cosx + 1 < 0 at this interval (2pi/3, 4pi/3) , so would that be my answer?

 

[cyby]

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...

 

[EL ALEM]

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?

 

[cyby]

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].

 

[EL ALEM]

Ok thanks.

 

[lasner12]

Ok so i found 2cosx + 1 < 0 at this interval (2pi/3, 4pi/3) , so would that be my answer?