## BloomLocke367 one year ago Solve this in interval form, from 0 to 2pi. Round to the nearest hundredth. $$-3sin2\theta=1.5$$

1. BloomLocke367

@mathmate @pooja195

2. BloomLocke367

@dan815 Will you come help?

3. BloomLocke367

can you help?

4. BloomLocke367

@pooja195

5. BloomLocke367

6. ganeshie8

as a start, divide -3 both sides

7. ganeshie8

$-3\sin(2\theta) = 1.5 \implies \sin(2\theta) = \frac{1.5}{-3} =-\frac{1}{2}$

8. ganeshie8

$\sin(2\theta) = -\frac{1}{2} \implies 2\theta = \sin^{-1}\left(-\frac{1}{2}\right)$ look up unit circle

9. BloomLocke367

I have the unit circle

10. BloomLocke367

what am I looking for?

11. ganeshie8

see for what angles the sin value is $$-\frac{1}{2}$$

12. BloomLocke367

sin is the y-value, right?

13. ganeshie8

|dw:1432912910233:dw|

14. BloomLocke367

so, 7pi/6 and 11pi/6

15. BloomLocke367

and I just round that to the nearest hundredth and I'll have my answer?

16. ganeshie8

Yes, $2\theta = \frac{7\pi}{6}+2k\pi,~ ~\frac{11\pi}{6}+2k\pi$ that means $\theta = \frac{7\pi}{12}+k\pi,~ ~\frac{11\pi}{12}+k\pi$

17. ganeshie8

but you want solutions in the interval 0 to 2pi, right ?

18. BloomLocke367

yes

19. ganeshie8

plugin k = 0, 1, 2 etc till you hit 2pi

20. BloomLocke367

why?

21. ganeshie8

$\theta = \frac{7\pi}{12}+k\pi,~ ~\frac{11\pi}{12}+k\pi$ plugin $$k=0$$ and get $\theta = \frac{7\pi}{12},~ ~\frac{11\pi}{12}$ plugin $$k=1$$ and get $\theta = \frac{7\pi}{12}+\pi,~ ~\frac{11\pi}{12}+\pi$

22. ganeshie8

because any integer value of $$k$$ gives an angle that is a solution to the given equation

23. ganeshie8

there are infinitely many solutions as there are infinitely many integers your question is about finding the solutions in the interval 0 to 2pi

24. BloomLocke367

okay.

25. BloomLocke367

and you plug in k=2, and get $$\large\frac{7\pi}{12}+2\pi,~\frac{11\pi}{12}+2\pi$$

26. ganeshie8

yes clearly thats more than $$2\pi$$, so we may stop now.

27. ganeshie8

k=0 and k=1 are enough

28. ganeshie8

find the corresponding $$\theta$$ values as you can see there are exactly four solutions in the interval 0 to 2pi

29. BloomLocke367

okay, and I just round it?

30. ganeshie8

the question asks you to round it so yeah

31. BloomLocke367

so, 1.83, 2.88, 4.97, and 6.02

32. ganeshie8

Looks good, you may double check with wolfram http://www.wolframalpha.com/input/?i=solve%20-3sin(2%5Ctheta)%3D1.5%2C%200%3C%5Ctheta%3C2pi&t=crmtb01

33. BloomLocke367

Okay, thanks!

34. ganeshie8

yw