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  1. Offline

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    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. Offline

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    n covers the scalar part....no need for 2(i.e. sub in n=2)......that's where you are going wrong
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    Why is there only one general solution instead of two?
  4. Offline

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    (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?
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    (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. Offline

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    (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. Offline

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    (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. Offline

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    Sorry boromir but I'm really not understanding you here

    Can anybody else give an explanation?
  9. Offline

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    (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. Offline

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    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. Offline

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    (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. Offline

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    (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. Offline

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    (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. Offline

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    (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. Offline

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    (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. Offline

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    Spoiler:
    Show
    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.
  17. Offline

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    (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. Offline

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    (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. Offline

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    (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...

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