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  1. pleasedtobeatyou's Avatar
    1 06-07-2012  20:18
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    General Solutions
    Finding the general solution for sin2x=0

    sin2x=0 2x = 0 2x=2n\pi+0 x=n\pi or 2x=(2n+1)\pi-0 2x=2n\pi+\pi x=n\pi+\pi/2

    The solution is nPi/2

    Where am I going wrong?
  2. boromir9111's Avatar
    2 06-07-2012  20:24
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    Re: General Solutions
    n covers the scalar part....no need for 2(i.e. sub in n=2)......that's where you are going wrong
  3. pleasedtobeatyou's Avatar
    3 06-07-2012  20:26
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    Re: General Solutions
    Why is there only one general solution instead of two?
  4. pleasedtobeatyou's Avatar
    4 06-07-2012  20:29
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    Re: General Solutions
    (Original post by boromir9111)
    n covers the scalar part....no need for 2(i.e. sub in n=2)......that's where you are going wrong
    I don't really understand what you're saying?
  5. boromir9111's Avatar
    5 06-07-2012  20:30
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    Re: General Solutions
    (Original post by pleasedtobeatyou)
    Why is there only one general solution instead of two?
    you are solving for when sinx = 0.....x = n*pi 'cause that's when it equals zero.....and that covers ALL the solutions, not just one.....to convince urself, sub in when n =2....that will equal zero....n =3 etc
  6. pleasedtobeatyou's Avatar
    6 06-07-2012  20:32
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    Re: General Solutions
    (Original post by boromir9111)
    you are solving for when sinx = 0.....x = n*pi 'cause that's when it equals zero.....and that covers ALL the solutions, not just one.....to convince urself, sub in when n =2....that will equal zero....n =3 etc
    But I'm solving for sin2x=0?
  7. boromir9111's Avatar
    7 06-07-2012  20:37
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    Re: General Solutions
    (Original post by pleasedtobeatyou)
    But I'm solving for sin2x=0?
    2x = n*pi
    x = (n*pi)/2

    so now we have sin(2*(n*pi)/2)....sub in for different values of n and you should see it gives various solutions
  8. pleasedtobeatyou's Avatar
    8 06-07-2012  20:42
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    Re: General Solutions
    Sorry boromir but I'm really not understanding you here

    Can anybody else give an explanation?
  9. boromir9111's Avatar
    9 06-07-2012  20:43
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    Re: General Solutions
    (Original post by pleasedtobeatyou)
    Sorry boromir but I'm really not understanding you here

    Can anybody else give an explanation?
    No need to be sorry

    What don't you get?
  10. pleasedtobeatyou's Avatar
    10 06-07-2012  20:51
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    Re: General Solutions
    Well, there are two formulae for finding the general solution for sin

    Equation 1: x=2n\pi+a Equation 2: x=(2n+1)\pi-a

    Solving sin2x=0 is 2x=0

    First equation: 2x=2n\pi x=n\pi

    Why would I not need a "2" in front of the n?

    And why is there only one general solution of n\pi/2

    Why is there not a second general solution from equation 2?
  11. electriic_ink's Avatar
    11 06-07-2012  20:58
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    Re: General Solutions
    (Original post by pleasedtobeatyou)
    Finding the general solution for sin2x=0

    sin2x=0 2x = 0 2x=2n\pi+0 x=n\pi or 2x=(2n+1)\pi-0 2x=2n\pi+\pi x=n\pi+\pi/2

    The solution is nPi/2

    Where am I going wrong?
    You're not. You just haven't realised that the two conditions x=n\pi and x=n\pi+\pi/2 can be simplified to x = nPi/2.
  12. pleasedtobeatyou's Avatar
    12 06-07-2012  21:03
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    Re: General Solutions
    (Original post by electriic_ink)
    You're not. You just haven't realised that the two conditions x=n\pi and x=n\pi+\pi/2 can be simplified to x = nPi/2.
    How would I do that? I'm struggling to find the link to simply n\pi/2
  13. electriic_ink's Avatar
    13 06-07-2012  21:12
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    Re: General Solutions
    (Original post by pleasedtobeatyou)
    How would I do that? I'm struggling to find the link to simply n\pi/2

    x=npi means that x = 0, pi, 2pi, ...
    x= npi + pi/2 means that x=pi/2, 3pi/2, 5pi/2, ...

    x= npi/2 means that x=0, pi/2, pi, 3pi/2, ...

    So they're the same.
  14. pleasedtobeatyou's Avatar
    14 06-07-2012  22:02
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    Re: General Solutions
    (Original post by electriic_ink)
    x=npi means that x = 0, pi, 2pi, ...
    x= npi + pi/2 means that x=pi/2, 3pi/2, 5pi/2, ...

    x= npi/2 means that x=0, pi/2, pi, 3pi/2, ...

    So they're the same.
    "x= npi + pi/2 means that x=pi/2, 3pi/2, 5pi/2"

    Looking back to sin2x=0,

    Wouldn't I be missing solutions? Because x could be pi and sin2(pi) would equal 0
  15. Lord of the Flies's Avatar
    15 07-07-2012  03:00
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    Re: General Solutions
    (Original post by pleasedtobeatyou)
    "x= npi + pi/2 means that x=pi/2, 3pi/2, 5pi/2"

    Looking back to sin2x=0,

    Wouldn't I be missing solutions? Because x could be pi and sin2(pi) would equal 0
    Just in case, don't forget that n\in\mathbb{Z}

    The idea is, you found two solutions which you can condense into one:

    x=n\pi=\{...-2\pi,-\pi,\;0,\;\pi,\;2\pi\;...\}

    x=n\pi+\pi/2=\{...-3\pi/2,-\pi/2,\;\pi/2,\;3\pi/2\;...\}

    Since both of these are true, you may condense them into: x=\{...-3\pi/2,-\pi,-\pi/2,\;0,\;\pi/2,\;\pi,\;3\pi/2\;...\}=n\pi/2

    So no, you aren't missing solutions.
  16. aznkid66's Avatar
    16 07-07-2012  05:33
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    Re: General Solutions
    Spoiler:
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    When you add on a " +2n\pi ", you're saying that the "cycle" (of solutions) repeats itself for every 2\pi. Therefore, the "cycle" (of solutions) must include all solutions within a range of 2\pi, for example all solutions from [0,2\pi) .

    (Original post by pleasedtobeatyou)
    sin2x=0 2x=2n\pi+0
    Thus, in the process of going from equation 1 to equation 2, you're saying that there is only one solution for \sin\theta=0 in a cycle of 0\leq\theta<2\pi. However, this is wrong. In addition to the solution \theta=0, there is the solution \theta=\pi. Since you are missing this solution (in all "cycles"), the two equations are not equivalent.

    So, to cover all the solutions, you must write:

    sin2x=0
    2x=0+2n\pi, pi+2n\pi

    Oh, whoops, my bad, you did that in the following... Right, just ignore this part then.


    (Original post by pleasedtobeatyou)
    x=n\pi+0
    (Original post by pleasedtobeatyou)
    x=n\pi+\pi/2

    The solution is nPi/2

    Where am I going wrong?
    The thing is, you're not going wrong. I think Lord of the Flies explained it quite clearly, so you might want to re-read his if you're still down here.

    If you have trouble making the logical step of combining two solution sets, I think it would be easier to look at the period. For example, the period of sin2x is pi/2, so solutions should be pi/2 apart, and that's what your resulting "cycle length" should end up as. I'm not sure it works though...

    Generally:

    \sin\theta=Q\Rightarrow\theta= $Arcsin$Q+2\pi k, \pi-$Arcsin$Q+2\pi k.

    While cosine is much nicer:

    \cos\phi=Q\Rightarrow\phi=\pm $Arccos$Q+2\pi k

    This can all be derived from the unit circle. Alternatively, use this Wikipedia article as a resource.
    Last edited by aznkid66; 08-07-2012 at 03:07.
  17. pleasedtobeatyou's Avatar
    17 07-07-2012  15:17
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    Re: General Solutions
    (Original post by Lord of the Flies)
    Just in case, don't forget that n\in\mathbb{Z}

    The idea is, you found two solutions which you can condense into one:

    x=n\pi=\{...-2\pi,-\pi,\;0,\;\pi,\;2\pi\;...\}

    x=n\pi+\pi/2=\{...-3\pi/2,-\pi/2,\;\pi/2,\;3\pi/2\;...\}

    Since both of these are true, you may condense them into: x=\{...-3\pi/2,-\pi,-\pi/2,\;0,\;\pi/2,\;\pi,\;3\pi/2\;...\}=n\pi/2

    So no, you aren't missing solutions.
    Thanks for clearing it up but is there any sort of method as such that I can use to condense the "2" general solutions?

    Or do I have to instinctively see it?
  18. Lord of the Flies's Avatar
    18 07-07-2012  15:59
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    Re: General Solutions
    (Original post by pleasedtobeatyou)
    Thanks for clearing it up but is there any sort of method as such that I can use to condense the "2" general solutions?

    Or do I have to instinctively see it?
    There is no method because you cannot always condense the two solutions.

    For instance:

    Spoiler:
    Show

    \sin x=\dfrac{1}{2} gives (n\in\mathbb{Z}):

    x_1=\dfrac{\pi}{6}+2n\pi

    x_2=\dfrac{5\pi}{6}+2n\pi


    Also, it isn't the end of the world if you don't see the simplification. Your solution is still correct if you give both solutions separately. So in your case you would have written:

    x=\{n\pi,\;\frac{\pi}{2}+n\pi\} which is fine. Condensing is just prettier and perhaps easier to visualise.
  19. Lord of the Flies's Avatar
    19 07-07-2012  16:34
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    Re: General Solutions
    (Original post by pleasedtobeatyou)
    Thanks for clearing it up but is there any sort of method as such that I can use to condense the "2" general solutions?

    Or do I have to instinctively see it?
    I am going to add something:

    If you end up needing to condense, (most of the time) it is that you didn't notice a simpler solution to start with. Take your equation, by picturing the circle you should be able to see that sin equals 0 with periodicity pi, not only 2pi. Essentially:

    \sin 2x=0\Rightarrow 2x=0+n\pi \Rightarrow x=\dfrac{n\pi}{2}

    Another example:

    \cos \dfrac{x}{7}=0\Rightarrow \dfrac{x}{7}= \dfrac{\pi}{2}+n\pi\Rightarrow x=\dfrac{7\pi}{2}+7n\pi

    Spoiler:
    Show

    Instead of going the long way:

    \cos \dfrac{x}{7}=0\Rightarrow \dfrac{x}{7}= \dfrac{\pi}{2}+2n\pi\Rightarrow x=\dfrac{7\pi}{2}+14n\pi or \dfrac{x}{7}= -\dfrac{\pi}{2}+2n\pi\Rightarrow x=-\dfrac{7\pi}{2}+14n\pi and then noticing you can condense these...
    Last edited by Lord of the Flies; 07-07-2012 at 18:40.
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