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Finding intervals of trig functions

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data
    solve the equation for x in the interval 0<=x<=2pi

    4cos(2x)+sin(x)=4

    3. The attempt at a solution
    I dont understand what the question is asking me to do? Where do I start and how can this equation be made into an appropriate equation so i can answer the question?
     
  2. jcsd
  3. Mar 21, 2010 #2

    rock.freak667

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    They want you to solve for x. But since sinx and cosx are periodic functions, you will an infinite number of solutions. So They gave you a specific interval.

    Start by trying to express cos(2x) in terms of sin(x)
     
  4. Mar 21, 2010 #3
    ok thanks
     
  5. Mar 22, 2010 #4
    so i simplified it to 4-2sin^2x+sin(x)=4 now what do i do, i dont really know what my aim is, as in what do i want my equation to look like before i start to find the values in the interval
     
  6. Mar 22, 2010 #5
    Be careful with your parenthesis.:tongue:
    [tex]cos(2x)=1-sin^{2}(x)[/tex] check your equation again :rolleyes:
     
  7. Mar 22, 2010 #6

    Mentallic

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    And then to make things more obvious, assign y=sin(x). You'll need to swap back later though.
     
  8. Mar 22, 2010 #7
    so 4sin^2(x)=sin(x) can be simplified to? im really bad at trig identities =(
     
  9. Mar 22, 2010 #8

    rock.freak667

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    right so bring the sinx on the left side and get

    4sin2x-sinx=0

    can you factor out sin(x) and get solutions?
     
  10. Mar 22, 2010 #9
    simplifying trig functions

    1. The problem statement, all variables and given/known data
    4cos(2x)+sin(x)=4

    in the interval 0<=x<=2pi

    3. The attempt at a solution
    i got this far

    sin(x)=8 sin^2(x)

    in the interval 0<=x<=2pi
    what is the question asking and how do i simplify this equation???
     
  11. Mar 22, 2010 #10

    Char. Limit

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    Re: simplifying trig functions

    Here's a clue...

    Divide by 8 sin(x)...
     
  12. Mar 22, 2010 #11

    danago

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    Re: simplifying trig functions

    Im guessing that the question is asking you to solve the equation for x, where x lies within the interval [tex]0 \leq x \leq 2\pi[/tex]

    You have simplified the equation which is good, but you now need to find x. Maybe write it as [tex]8 sin^2 (x) - sin(x) = 0[/tex]. How would you solve it from there? HINT: Try factorizing!
     
    Last edited: Mar 22, 2010
  13. Mar 22, 2010 #12

    danago

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    Re: simplifying trig functions

    You need to be careful with this though, because what if [tex]sin(x)=0[/tex]?

    Purely by inspection of the equation, i can see that x = 0 is a possible solution. If you divide everything by [tex]8 sin(x)[/tex] however, the equation becomes [tex]0.125 = sin(x)[/tex], but now x=0 is no longer a solution.
     
    Last edited: Mar 22, 2010
  14. Mar 22, 2010 #13

    Char. Limit

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    Re: simplifying trig functions

    Just remember the x=0 solution... it's not that hard, really.
     
  15. Mar 23, 2010 #14
    Re: simplifying trig functions

    would the factorized version look like this???
    (8sin(x)+1)(sin(x)-1)=0
     
  16. Mar 23, 2010 #15

    danago

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    Re: simplifying trig functions

    Im not really sure how you came to that result. If you try to expand it out again, you dont get back to
    [tex]
    8 sin^2 (x) - sin(x) = 0
    [/tex]

    The way to factorize it would be to note that sin(x) is a common factor of both terms:

    [tex]8 sin^2 (x) - sin(x) = sin(x)(8 sin (x) - 1) = 0[/tex]
     
  17. Mar 23, 2010 #16

    Mentallic

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    :devil:
     
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