1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Easy? Trignometry question

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the following equations or inequalities in the interval [0, 2pi).

    2sin2x - 5sinx + 3 < 0


    2. Relevant equations



    3. The attempt at a solution

    (2sin2x - 3)(sinx - 1)
    sinx = 3/2 or sinx = 1

    Not sure what to do now

    sinx = 3/2 is impossible?
    sinx = 1 then x = 90 = pi/2
     
  2. jcsd
  3. Sep 27, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    just correcting your square...

    2sin2x - 5sinx + 3 = (2sinx-3)(sinx-1)

    However rather than finding the bounding points (=0, though your thinking was correct) you have an inequality
    (2sinx-3)(sinx-1) < 0

    so what conditions need to be satisfied for (2sinx-3)(sinx-1) to be negative?
     
  4. Sep 28, 2009 #3
    Do I need to sub in points to see rather it is positive or negative before and after sinx - 1?
     
  5. Sep 28, 2009 #4

    Mark44

    Staff: Mentor

    Your inequality in factored form is (2sinx-3)(sinx-1) < 0.
    As you have already noticed, the first factor can't be zero, which means that it is either always positive or always negative, no matter what x value you substitute. Determine which of these it is.

    For the product of the two factors to be negative, they have to be opposite in sign.
     
  6. Sep 28, 2009 #5

    lanedance

    User Avatar
    Homework Helper

    following on from what Mark said, if the sin's are confusing, first solve for y, ie
    (2y-3)(y-1)<0

    then translate that to the original problem, by y = sinx, knowing that only y values in the range [-1,1] are allowable solutions for x
     
  7. Sep 28, 2009 #6
    So I think that (2sinx-3) is always neagative, this means that I need to find values of x where (sinx -1) is positive to satisfy the inequality?
     
  8. Sep 28, 2009 #7

    lanedance

    User Avatar
    Homework Helper

    sounds like you're on the right track to me
     
  9. Sep 28, 2009 #8
    Does this mean I need to solve sinx > 1?
    I don't think there are any values where sinx > 1. Or am I confused about what I'm doing?
    sinx is 1 when x is 90
     
  10. Sep 28, 2009 #9

    Mark44

    Staff: Mentor

    sin(x) = 1 when x = (90 + k*360) degrees, but are there any values of x for which sin(x) > 1? If you're not sure, see lanedance's post 5.
     
  11. Sep 29, 2009 #10
    What happens if there are no values where sinx > 1?
     
  12. Sep 29, 2009 #11
    Then the inequality has no solutions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Easy? Trignometry question
  1. Obstuse trignometry (Replies: 8)

  2. Easy Question! (Replies: 4)

  3. Basic trignometry (Replies: 8)

Loading...