Results are out! Find what you need...fast. Get quick advice or join the chat
Hey! Sign in to get help with your study questionsNew here? Join for free to post

General Solutions

Announcements Posted on
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    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?
    • 36 followers
    Offline

    ReputationRep:
    n covers the scalar part....no need for 2(i.e. sub in n=2)......that's where you are going wrong
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    Why is there only one general solution instead of two?
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    (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?
    • 36 followers
    Offline

    ReputationRep:
    (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
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    (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?
    • 36 followers
    Offline

    ReputationRep:
    (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
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    Sorry boromir but I'm really not understanding you here

    Can anybody else give an explanation?
    • 36 followers
    Offline

    ReputationRep:
    (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?
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    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?
    • 3 followers
    Offline

    ReputationRep:
    (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.
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    (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
    • 3 followers
    Offline

    ReputationRep:
    (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.
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    (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
    • 19 followers
    Offline

    ReputationRep:
    (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.
    • 3 followers
    Offline

    ReputationRep:
    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.
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    (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?
    • 19 followers
    Offline

    ReputationRep:
    (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 followers
    Offline

    ReputationRep:
    (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...

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: July 7, 2012
New on TSR

The future of apprenticeships

Join the discussion in the apprenticeships hub!

Article updates
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.