## anonymous one year ago Solve: 1-sin(2πt)=0, for 0≤t≤2.

1. Michele_Laino

we have: $\Large \sin \left( {2\pi t} \right) = 1$

2. Michele_Laino

now: sin x= 1 when: $x = \frac{\pi }{2},2\pi + \frac{\pi }{2},4\pi + \frac{\pi }{2}...$

3. Michele_Laino

for example, for the first solution, pi/2, we can write: $2\pi t = \frac{\pi }{2}$ please solve for t

4. anonymous

t=2π?

5. triciaal

no try again

6. anonymous

t=π^2?

7. triciaal

|dw:1437338469566:dw|

8. triciaal

|dw:1437338527982:dw|

9. triciaal

need the words?

10. triciaal

are you here?

11. anonymous

Oh... lol I am an idiot. xD

12. Nnesha

nah u r no8 :=)

13. anonymous

Wait, so is 1/4 the answer?

14. zepdrix

$\large\rm 2\pi t=\frac{\pi}{2}+2\pi k,\qquad k\in \mathbb{Z}$Dividing by 2pi,$\large\rm t=\frac{1}{4}+k,\qquad k\in \mathbb Z$When k=0, yes you get t=1/4. That is one of your solutions. Think of other integers though. Which other integers for k will keep our t between 0 and 2?

15. anonymous

Um... 3pie over 2?

16. freckles

integers. 3pi/2 is not an integer t=1/4+k where k is an integer are your candidates for solutions in the interval [0,2] notice for k=0 you have t=1/4 what happens if you let k=1?

17. anonymous

So... 0 and 2?

18. freckles

so we want to find t in [0,2] such that $0 \le t \le 2 \\ 0 \le \frac{1}{4}+k \le 2 \text{ from here you decide what \to \choose for } k \\ 0-\frac{1}{4} \le k \le 2-\frac{1}{4} \\ \text{ I subtracted } \frac{1}{4} \text{ on both sides } \\ \frac{-1}{4} \le k \le \frac{7}{4}$ what are the only integers in that inequality ?

19. anonymous

0 and 2?

20. freckles

do you mean 0 and 1?

21. freckles

7/4 is less than 2 so 2 can't be included in that inequality above

22. anonymous

Oh i see.

23. freckles

|dw:1437428007076:dw| only integers between -1/4 and 7/4 is 0 and 1

24. freckles

so these are the integers that @zepdrix was referring to

25. freckles

he already entered in 0 for k to get one solution for t now you just have to enter in 1 for k to get the other solution for t

26. freckles

$t=\frac{1}{4}+k \\ \text{ when } k=0 \text{ we have } t=\frac{1}{4}+0=\frac{1}{4} \\ \text{ when } k=1 \text{ we have } t=?$

27. freckles

I'm just asking you to replace k with 1

28. freckles

t=1/4+1=?

29. anonymous

5/4

30. freckles

yes t=1/4 or t=5/4 are the only solutions for t in the given interval

31. anonymous

Wow. that was intense... lol

32. anonymous

Thank you

33. freckles

np lol the word intense is intense by itself (to me anyways)

34. anonymous

Haha. Yea it is huh? xD