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# General Solutions

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1. Finding the general solution for sin2x=0

The solution is nPi/2

Where am I going wrong?
2. n covers the scalar part....no need for 2(i.e. sub in n=2)......that's where you are going wrong
3. Why is there only one general solution instead of two?
4. (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?
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. (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?
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. Sorry boromir but I'm really not understanding you here

Can anybody else give an explanation?
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. Well, there are two formulae for finding the general solution for sin

Solving sin2x=0 is 2x=0

First equation:

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

And why is there only one general solution of

Why is there not a second general solution from equation 2?
Finding the general solution for sin2x=0

The solution is nPi/2

Where am I going wrong?
You're not. You just haven't realised that the two conditions and can be simplified to x = nPi/2.
12. (Original post by electriic_ink)
You're not. You just haven't realised that the two conditions and can be simplified to x = nPi/2.
How would I do that? I'm struggling to find the link to simply
How would I do that? I'm struggling to find the link to simply

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. (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
"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

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

Since both of these are true, you may condense them into:

So no, you aren't missing solutions.
16. Spoiler:
Show
When you add on a " ", you're saying that the "cycle" (of solutions) repeats itself for every . Therefore, the "cycle" (of solutions) must include all solutions within a range of , for example all solutions from .

Thus, in the process of going from equation 1 to equation 2, you're saying that there is only one solution for in a cycle of . However, this is wrong. In addition to the solution , there is the solution . Since you are missing this solution (in all "cycles"), the two equations are not equivalent.

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

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

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:

.

While cosine is much nicer:

This can all be derived from the unit circle. Alternatively, use this Wikipedia article as a resource.
17. (Original post by Lord of the Flies)
Just in case, don't forget that

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

Since both of these are true, you may condense them into:

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?
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

gives

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:

which is fine. Condensing is just prettier and perhaps easier to visualise.
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:

Another example:

Spoiler:
Show

Instead of going the long way:

or and then noticing you can condense these...

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