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Trig. Equation

  1. Mar 19, 2014 #1
    1. The problem statement, all variables and given/known data

    (sin(2x)) * (2sin(2x)-1)=0

    2. The attempt at a solution

    sin (2x) = 0
    sinx=0

    x= 0, π

    2sin(2x)-1=0
    sin (2x) = 1/2
    sinx = 1/4
    x = π/12 ; 11π/12

    There are two more solutions but I cannot seem to find them.
     
  2. jcsd
  3. Mar 19, 2014 #2

    HallsofIvy

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    Are you to find x in the interval [itex]0\le x< 2\pi[/itex]? It would be better if you would tell us that!
    Yes, sin(x)= 0 for x= 0 and [itex]x= \pi[/itex].

    For 2 sin(2x)- 1= 0, yes, that leads to sin(2x)= 1/2.
    But you cannot then declare that sin(x)= 1/4!!
    The "2" is inside the function- you cannot divide by 2 until after you have removed the sine.
    (That's a howler of an error! I really hope that was carelessness.)

    From sin(2x)= 1/2 you get [itex]2x= \pi/6, \pi= \pi/6= 5\pi/6, 2\pi+ \pi/6= 13\pi/6, 3\pi- \pi/6= 17\pi/6[/itex]

    Notice that I have gone to 0 to [itex]2\pi[/itex] because I am going to divide by 2:

    [itex]x= \pi/12[/itex], [itex]x= 5\pi/12[/itex], [itex]13\pi/12[/itex], [itex]17\pi/12[/itex].
     
  4. Mar 19, 2014 #3
    My teacher had told me that I could remove the two from sin(2x) right away... sorry about that.

    From the sin(2x) = pi/6 ... i am kind of lost

    I know how you got pi/6, but the rest, I don't quite understand.
     
  5. Mar 19, 2014 #4

    Ray Vickson

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    I hope that what your teacher means is that if you have sin(2x) = 0, then setting y = 2x you have sin(y) = 0, NOT sin(x) = 0. In fact, if sin(2x) = 0 then sin(x) is as far from 0 as the 'sin' can get: we would have either sin(x) = +1 or sin(x) = -1.

    Also, what you wrote above makes no sense: you do not have sin(2x) = pi/6; you do have sin(2x) = 1/2 = sin(pi/6), so 2x = pi/6.
     
    Last edited: Mar 19, 2014
  6. Mar 19, 2014 #5
    Yes, I meant to write 2x = pi/6 not sin 2x = pi/6
     
  7. Mar 19, 2014 #6
    I mainly don't understand the part where HallsOfIvy wrote pi = pi/6 = 5pi / 6 ...

    Where did he get the pi = pi/6 = 5pi / 6 ... from?
     
  8. Mar 19, 2014 #7

    Ray Vickson

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    That is NOT what he wrote. Read it again.

    He is just listing the several possible values of x, so what he wrote is shorthand for x = 0 or x = pi/6 or x = 5pi/6 or .... . He is not saying those values are equal to each other.
     
  9. Mar 19, 2014 #8

    SammyS

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    I believe Halls meant to write

    From sin(2x)= 1/2 you get [itex]2x= \pi/6,\ \pi- \pi/6= 5\pi/6,\ 2\pi+ \pi/6= 13\pi/6,\ 3\pi- \pi/6= 17\pi/6\ .[/itex]

    in particular, [itex]\ ... \pi- \pi/6= 5\pi/6,\ ...[/itex]
     
  10. Mar 20, 2014 #9
    Yes. I have understood the problem.
     
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