## anonymous one year ago What values for,(posted below) satisfy the equation?

1. anonymous

This fills in the blank

2. anonymous

This follows the question

3. anonymous

@UsukiDoll

4. UsukiDoll

I don't understand what the question needs besides from 0< x< 2pi means from 0 to 360 degrees which is the full circle

5. UsukiDoll

there is a trig identity $\tan^2(\theta) = 1-\sec^2(\theta)$

6. anonymous

What values for, |dw:1436466383880:dw| satisfy the equation?

7. UsukiDoll

maybe we can replace that tan with what I've wrote... ??? dah I've never seen this formatted question before x.x |dw:1436502474870:dw|

8. anonymous

The rest of the equation is in the second attachment at the top

9. anonymous

|dw:1436466531453:dw|

10. UsukiDoll

this is what I'm thinking but I'm not 100% sure ... use the identity on the left hand side $\tan^2(\theta) = 1-\sec^2(\theta)$

11. UsukiDoll

and then add $\frac{3}{2}\sec(\theta)$ to both sides so we end up with some sort of quadratic equation (well similar )

12. Oleg3321

ehhh im not so sure how to do this one either.

13. anonymous

|dw:1436466790578:dw|

14. anonymous

15. misty1212

HI do what was suggested above write it as a quadratic equation in terms of secant then solve

16. anonymous

I don't know how to do these equations at all

17. misty1212

lol then cheat!

18. anonymous

haha well i was thinking you guys could give me the right answer cuz i cant find it

19. misty1212

i can show you how to find it if you like it will take a bit of time http://www.wolframalpha.com/input/?i=tan%5E2%28x%29%3D-3%2F2sec%28x%29

20. UsukiDoll

yes @misty1212 we write it as a quadratic equation in terms of secant and then solve

21. anonymous

@misty1212 anything can help, i just need the right answer

22. UsukiDoll

I know most of the process... but towards the end . I might crash

23. UsukiDoll

I think it has something to do with adding 2pi or subtracting 2pi ...hence more than one answer

24. UsukiDoll

|dw:1436503293700:dw|

25. UsukiDoll

|dw:1436503337897:dw|

26. UsukiDoll

|dw:1436503367107:dw| now multiply the entire equation by -2 to get rid of the fraction and the negative

27. anonymous

idk, i dont have a calculator at the moment sadly lol

28. UsukiDoll

|dw:1436503536319:dw|

29. UsukiDoll

|dw:1436503556177:dw|

30. UsukiDoll

now let all secants be x for now because that quadratic equation looks mean

31. UsukiDoll

|dw:1436503613776:dw|

32. UsukiDoll

oh yes we can factor this woo!

33. anonymous

ok im learning lol

34. UsukiDoll

|dw:1436503685977:dw|

35. UsukiDoll

I'm using b^2-4ac so I would know what that part is already for that quadratic formula

36. UsukiDoll

|dw:1436503730219:dw|

37. UsukiDoll

|dw:1436503768417:dw|

38. UsukiDoll

|dw:1436503804424:dw|

39. anonymous

So how do we find one of my answers from A-D? lol

40. UsukiDoll

$(\sec(\theta)-2)(\sec(\theta)+\frac{1}{2})=0$ something is screwed up... I am expanding and the last term isn't working out

41. anonymous

@satellite73 sos

42. UsukiDoll

nice I found the problem -_-! only one of them is right

43. UsukiDoll

but strange taking the quadratic equation approach always works...

44. anonymous

Can you explain, or do you just wanna give the answer?

45. UsukiDoll

-__- *facepalm* I made that error by not getting rid of that fraction .. argh.. ok I thought we were done for.

46. UsukiDoll

$(\sec(\theta)-2)(2\sec(\theta)+1)=0$

47. UsukiDoll

so now we have to solve this separately . . . which I swear we have already done it when I found my roots.

48. anonymous

Alrightyyy

49. UsukiDoll

$\sec(\theta)-2=0$ adding 2 to both sides $\sec(\theta) =2$ $(2\sec(\theta)+1)=0$ that's $\sec(\theta) = \frac{-1}{2}$

50. UsukiDoll

turns out that we already did this step and not even know it ^ because I've used the quadratic equation to get its roots.

51. UsukiDoll

so secant = 1/cosine this would be a bit easier to work with

52. anonymous

ok whats next

53. UsukiDoll

so my cosine has to be $\frac{1}{2}$ for $\sec(\theta) =2$ and to check secant of 1/2 = 1/(1/2) = 2

54. UsukiDoll

|dw:1436504658208:dw| based on that triangle cosine 1/2 is 60 degrees otherwise known as $\frac{\pi}{3}$ in radians

55. anonymous

56. UsukiDoll

yeah, but I want to get the other answer from that choice.

57. UsukiDoll

we got our pi/3. let's see if I can grab 5pi/3

58. UsukiDoll

which is 300 degrees.

59. UsukiDoll

OH MAN! I am making this hard on myself again ROFL! We just have to do this $2\pi-\frac{\pi}{3}$ because according to our restriction we are in the unit circle 0< theta < 360 means from 0 to 360.. since we need 300 degrees we are in the fourth quadrant.

60. UsukiDoll

$\frac{3}{3} \times 2\pi-\frac{\pi}{3} = \frac{6 \pi-\pi}{3}$

61. UsukiDoll

$\frac{5\pi}{3}$

62. UsukiDoll

there. XDDDDDDDDDDDDDDDDDDDDDDD

63. UsukiDoll

I can't believe I just did that ... it's been 6 years....

64. anonymous

yay you did it lol

65. UsukiDoll

the wording on the question threw me off... because in my textbooks it was a bit simpler so I had to think it through for a bit.

66. anonymous

yeah its tough sometimes, but at least you guys are great help

67. UsukiDoll

thanks :D

68. anonymous

you're welcome. i gave you a medal