1. I would simply get a=b ?

it would only be at rest when a=b?

2. No, that is not the only solution, though it is certainly one solution. Not the most general answer, though. When do two sine functions equal each other? What is true of the arguments of the sine functions?

3. generally you would need to just add pi ?

4. This is what I do..

sin(2t-2a)=sin(2t-2b)

take sin inverse of both side gives

2t-2a=2t-2b a=b ... but this also satisfies when I add pi

eg.. sin(2t-2a+pi)=sin(2t-2b+pi) ??

5. Not pi but... what?

6. I'm confused... are you think about any revolution ?

npi ?

7. No, you're on the right track. But the sin function is not pi-periodic. What is the period of sin?

8. oh 2pi...

I was thinking of the second quadrant...

but what does this have to do with it?

I can't see the connection?

9. So, if a function is 2pi-periodic, that means sin(t+2 pi n) = sin(t) for all t. Can you use this information in the problem?

10. I don't know.. I'm a little confused

I know I'm trying to get rid of t, and trying to get a relationship that is always greater than b.

But I totally lost it.

11. Well, I hate to be tough on you, but I think it's the best thing for you. I feel that I've given you enough hints to finish the problem. What's left really isn't all that earth-shattering. Just look over the hints, maybe collect everything in one place, compare equations, and see if something doesn't come to mind.

12. I thank you for your time and effort, and yes I'm trying but I'm just wondering because I started off with trying to make

d(t-a) =/= 0 for t>b

I'll ask my tutor, but thanks once again

Thanks once again!

13. You're welcome. Good luck!

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