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!

Solving trig equations involving aperiodic functions

  1. Jul 14, 2014 #1
    1. The problem statement, all variables and given/known data

    kcfVccY.png

    2. Relevant equations



    3. The attempt at a solution


    C)

    Principal domain:

    [tex]-\dfrac{\pi}{2}\leq 2x^2+x-1\leq \dfrac{\pi}{2}[/tex]

    [tex]-1.41099 \leq x \leq 0.91099[/tex]

    Next:

    [tex]\sin(2x^2+x-1)=\dfrac{2}{5}[/tex]

    [tex]2x^2+x-1=\arcsin(\dfrac{2}{5})[/tex]

    [tex]x_1=-1.1265; x_2=0.62650[/tex]

    Since [tex]\sin(\theta)=-\sin(\theta)+\pi[/tex] we get:

    [tex]\sin(-(2x^2+x-1)+\pi)=\dfrac{2}{5}[/tex]

    [tex]-2x^2-x+1+\pi=\arcsin(\dfrac{2}{5})[/tex]

    Hence:

    [tex]x_3=1.1384; x_4=-1.6384[/tex]

    Although, I wouldn't be able to find any more values of x by the methods I have been introduced to, since the functions is aperiodic.


    By the same method I can find two values of x in D), but that's it. Although, I think they want me to do something else, rather than values for x.



    P.S. Also, I haven't been introduced to solving trig equations with aperiodic functions, so it must be that they want to stretch me a little. So the tasks are probably rigged somehow, like in the case of C) I can find exactly four values with the methods I've been shown (well, as I understand), and that's what they want from me in the problem.

    So, I guess it must be similar with the D) task, but I can't figure it out.
     
    Last edited: Jul 14, 2014
  2. jcsd
  3. Jul 14, 2014 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    If ##\sin(a) = \frac{2}{5}##, what are all the possible values for ##a##?
     
  4. Jul 14, 2014 #3

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Where does this come from? Why must ##x## satisfy this? This isn't given in the question.
     
  5. Jul 14, 2014 #4
    It's just a procedure from my book to find 1 value (well, in this case there are two, because this problem involves a quadratic function) on this restricted domain and then work from it deriving general solutions.

    Well, in this case
    ## a=-(\arcsin(\dfrac{2}{5})-\pi) ##
     
    Last edited: Jul 14, 2014
  6. Jul 14, 2014 #5
    Oh wait, so I guess then:

    ## 2x^2+x-1=-(\arcsin(\dfrac{2}{5})-\pi) ## ?
     
  7. Jul 14, 2014 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Those are not all the possible values for ##a##, but ok, setting

    [tex]2x^2 + x -1 = -(\arcsin(\dfrac{2}{5})-\pi)[/tex]

    should work.
     
  8. Jul 14, 2014 #7
    Yes, I fixed it, forgot to include something:

    ## a=-(\arcsin(\dfrac{2}{5})-\pi +2k \pi), k \in \mathbb Z ##
     
    Last edited: Jul 14, 2014
  9. Jul 14, 2014 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Those are still not all the solutions, for example ##\arcsin(2/5)## isn't included in there.
     
  10. Jul 14, 2014 #9
    Right... Silly me. But I get the idea.

    ## a=\arcsin(\dfrac{2}{5})+2k \pi, k \in \mathbb Z ##


    ## a=-\arcsin(\dfrac{2}{5})+\pi +2k \pi, k \in \mathbb Z ##
     
    Last edited: Jul 14, 2014
  11. Jul 14, 2014 #10

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Right, and all individual values for k give us a solution. So this can be used to find many solutions of your original equation.
     
  12. Jul 14, 2014 #11
    Oh, actually I just noted that ## k \notin \mathbb Z## but rather ## k \in \mathbb W ##. Otherwise the solutions are complex numbers. And any ##k \in \mathbb W## will give both negative and positive solutions.

    Has to do with the "quadraticity" of the function.
     
  13. Jul 14, 2014 #12

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What is ##\mathbb W \ ?##
     
  14. Jul 14, 2014 #13
    Whole numbers. I wasn't sure if you could denote it like that, but I did :).

    Well, I guess it actually isn't allowed, so then it should be ## k \in \mathbb N, k=0 ##

    Or maybe like this http://mathworld.wolfram.com/Z-Star.html

    So ## k \in \mathbb{Z}^+ ##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Solving trig equations involving aperiodic functions
  1. Solving trig functions (Replies: 1)

  2. Solving Trig Equations (Replies: 11)

  3. Solving trig equations (Replies: 14)

Loading...