Trig #3 (Part solutions included)

[lovemake1]

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.

 

[vela]

Sounds good.

 

[HallsofIvy]

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.

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.

secx= 1 is the same as 1/cosx= 1 or cos x= 1. sec x= 2 is the same as cos(x)= 1/2. and note that 1< 1/cos(x)< 2 is the same as cos(x)< 1< 2cos(x) or 1/2< cos(x)< 1.
Of course, if you are asked for all such x, not just between 0 and \pi/2 or 0 and 2\pi, there will be many intervals in which that is true.

 

[lovemake1]

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

 

[eumyang]

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

No, this is not correct. You're forgetting that you're supposed to solve an inequality, not an equation. Reread HallsofIvy's post.

 

[lovemake1]

oh right, here are my fixed answer.

x belongs to (1pi/3, 1pi/3)

are they correct?
i didnt include 0, because x must be < cos = 1