The discussion revolves around solving the inequality sec²x - 3secx + 2 < 0, which involves understanding the behavior of the secant function and its relationship to cosine. Participants are exploring the implications of the factored form of the inequality and the intervals where secx lies between 1 and 2.
The conversation is ongoing, with some participants providing feedback on each other's interpretations and attempts. There is a recognition of the need to clarify the distinction between solving an inequality and an equation, and some guidance has been offered regarding the correct approach to finding solutions.
Participants are working within the constraints of the interval [0, 2π) and are considering the implications of the inequality on the values of x. There is an acknowledgment of the potential for multiple solutions across different intervals.
ab< 0 as long as a and b have different signs. Since 2> 1, secx- 2< secx- 1 so you want sec x- 2< 0 and sec x- 1> 0. That is, as you say, 1< sec x< 2.lovemake1 said:Homework Statement
sec^2x - 3secx + 2 < 0
Homework Equations
The Attempt at a Solution
sec^2x - 3secx + 2 < 0
ok so this can be written as, (secx - 2)(secx -1) < 0
this means secx is between 1 and 2 right ?
secx = 2 and secx = 1
would we find all the solutions that's between 1 and 2 ?
please check if my understanding is correct.
No, this is not correct. You're forgetting that you're supposed to solve an inequality, not an equation. Reread HallsofIvy's post.lovemake1 said:so the two solutions are, cosx = 1 and cosx = 1/2
since x belongs to [0,2pi), x = 0, 1/3pi
Could someone correct me if I am wrong, i think i got the right answer.
please check thanks