## anonymous one year ago What values for theta(0< or equal to theta < or equal to 2pi) satisfy the equation? 3sin theta = sin theta - 1 tan^2 theta = -3/2sec theta Help! I have no idea what im doing

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1. campbell_st

for the 1st equation $3\sin(\theta) = \sin(\theta) - 1$ so start by subtracting $\sin(\theta)$ from both sides of the equation... what do you get..?

2. anonymous

ummm 4 sin=-1?

3. anonymous

or 2sin=-1?

4. campbell_st

its the 2nd option $2 \sin(\theta) =-1$ if you divide both sides of the equation by 2 you get $\sin(\theta) = -\frac{1}{2}$ does that make sense..?

5. anonymous

yes... so 2 would be tan^2 = -3/2 sec sin/cos ^2 = -3/2 1/cos = sin^2/cos=-3/2 ?

6. campbell_st

well getting back to the 1st equation... you can now solve this... find sin = 1/2 and sin is negative in the 3rd and 4th quadrants so the answer is 180 + theta and 360 - theta hope that is ok..

7. anonymous

pi/3 and 5pi/3

8. anonymous

??

9. campbell_st

for the 2nd equation rewrite it as $\frac{\sin^2(\theta)}{\cos^2(\theta)} = -\frac{3}{2\cos(\theta)}$ multiply both sides by $\cos^2(\theta)$ and you get $\sin^2(\theta) = \frac{-3\cos(\theta)}{2}$ then using sin^2 + cos^2 = 1 you can make the substitution $1 - \cos^2(\theta) = -\frac{3\cos(\theta)}{2}$ which becomes $\cos^2(\theta) - \frac{3\cos(\theta)}{2} - 1 = 0$ so you can solve the quadratic hope it makes sense

10. campbell_st

ok... so if you are working in radians the angles are $\pi + \theta~~and~~~2\pi - \theta$

11. anonymous

Now you lost me completely.

12. campbell_st

remember from an exact value triangle $\sin(\theta) = \frac{1}{2}~~~then~~~~\theta = \frac{\pi}{6}$

13. anonymous

But there must also be a second one... so pi/6 and 3pi/6?

14. campbell_st

here is the diagram |dw:1433489868047:dw| since sin is negative it means the angles are in the 3rd and 4th quadrants... so the 3rd quadrant angle is $\pi + \frac{\pi}{6}$ and the 4th quadrant angle is $2\pi - \theta$

15. campbell_st

oops 4th quadrant is $2\pi - \frac{\pi}{6}$

16. anonymous

Still confused. It cant be negative....

17. anonymous

pi/6?!?!?!?

18. anonymous

and 2pi/6 = pi/3???

19. campbell_st

yes it can here is a graph of the sin curve |dw:1433490149420:dw|

20. anonymous

none of my choices are negative though,

21. campbell_st

that is correct.... but some of you choices may include angles that are larger than pi and smaller than 2pi... is that correct...?

22. anonymous

this is still for the first one?

23. campbell_st

angles between pi and 2 pi... return negative values... for sin e.g. find sin(3pi/2) on your calculator... it will return a value of -1

24. anonymous

pi/3, 5pi/3 2pi/3. 4pi/3 pi/6, 3pi,6 7pi/6, 11pi/6

25. anonymous

so like d?

26. campbell_st

yes... we are still working on the 1st equation...

27. campbell_st

so like.... that's my guess

28. campbell_st

put it into your calculator does sin(7pi/6) = -1/2 and does sin(11pi/6) = -1/2 try it

29. anonymous

I can use the unit circle too, yes?

30. anonymous

if so, it is yes for both

31. campbell_st

well there you go... sin can be negative... and the angles are larger then pi radians...

32. anonymous

so thats the answer? no more steps?

33. campbell_st

well as I said... that's my best guess... and you told me the unit circle agreed...

34. anonymous

so for 2... tan^2 = -3/2sec (sin/cos )^2= -3/2 1/cos ??

35. campbell_st

the unit circle produces the sin curve... that's important

36. anonymous

trig is not my best section...

37. campbell_st

well I'd rewrite 2 $\frac{\sin^2(\theta)}{\cos^2(\theta)} = \frac{-3}{2\cos(\theta)}$ does that make sense..?

38. campbell_st

then multiply both sides by cos^2$\sin^2(\theta) = \frac{-3 \cos(\theta)}{2}$ is that ok..?

39. anonymous

yes.. and then i square root everything?

40. campbell_st

no, make a substitution sin^2 = 1 - cos^2 this come from sin^2 + cos^2 = 1 so you get $1 - \cos^2(\theta) = \frac{-3\cos(\theta)}{2}$ is that ok

41. anonymous

okay.. so cross cancel the cos?

42. campbell_st

no re-write the equation as $\cos^2(\theta) - \frac{3}{2} \cos(\theta) - 1 = 0$ so you have a quadratic equation you need to solve...

43. anonymous

how do i set it up

44. campbell_st

my advice is to use the general quadratic formula $\cos(\theta) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

45. anonymous

but how since theres more than 1?

46. campbell_st

well the coefficients of the quadratic are a = 1, b = -3/2 and c = -1 if you struggled with trig you must be good at solving quadratics

47. campbell_st

then only consider the value that is less than 1... a solution that is greater than 1 is undefined

48. campbell_st

again, you need to consider where cos is positive... 1st and 4th quadrants... so you will have 2 solutions...

49. campbell_st

50. anonymous

3.2+/- sqrt -3/2^2+4 ----- 2 so 4.95 and 1.45?

51. Jack1

how do u convert from radians to degrees again...?