## anonymous 5 years ago which value of theta satisfies the equation 2cos^2theta - cos theta=0

1. anonymous

$cos^2(\theta)-cos(\theta)=0$ $cos(\theta)(cos(\theta)-1)=0$ $cos(\theta)=0$ or $cos(\theta)=1$

2. anonymous

oooooooops forgot the two!

3. anonymous

ignore my post, or just modify with the two.

4. anonymous

is there a formula to memorize forthis?

5. anonymous

$cos(\theta)(2cos(\theta)-1)=0$ $cos(\theta)=0$ $cos(\theta)=\frac{1}{2}$

6. anonymous

oh no the first part has nothing at all to do with trig

7. myininaya

costheta(2costheta-1)=0 cost=0 => t=pi/2,3pi/2,... 2npi+pi/2 and 2npi+3p/2 for n=0,1,2,... 2cost-1=0 => cost=1/2 => t=pi/3,11pi/6,.... 2npi+pi/3 and 2npi+11pi/6 for n=0,1,2..

8. anonymous

like solving $2x^2-x=0$ $x(2x-1)=0$ $x=0$ $x=\frac{1}{2}$

9. anonymous

oh no im confused

10. anonymous

the trig part comes now.

11. myininaya

oops that 11pi/6 is wrong its 5pi/3

12. anonymous

the first job is to solve for $cos(\theta)$ the next step is to solve for $\theta$

13. anonymous

the first steps have nothing at all to do with trig. no formulas, no nothing.

14. anonymous

ok

15. anonymous

think of $cos(\theta)$ as x and solve for x. you have $2x^2-x=0$ and this is easy to solve

16. anonymous

we get $x=0$ or $x=\frac{1}{2}$ so now $cos(\theta)=0$ or $cos(\theta)=\frac{1}{2}$

17. anonymous

the trig part comes now, seeing if we can find numbers for theta that will make this work.

18. anonymous

can you do the next step, that is, can you find theta?

19. anonymous

im not sure/

20. anonymous

isnt it 0?

21. anonymous

ok so first off we want to find where $cos(\theta)=0$

22. anonymous
23. myininaya

just look between 0 and 2pi first to see where cosine is 0

24. anonymous

best cheat sheet. look on it to see where cosine is 0. it is not at 0, because $cos(0)=1$

25. myininaya

then after just add 2npi for n=0,1,2,3,... blah blah

26. anonymous

cos is the x correct?

27. anonymous

yes we already know x=0, so therefore $cos(\theta)=0$ now we are looking for $\theta$ that is we are looking for an angle (number) whose cosine is 0

28. myininaya

what happens at pi/2?

29. anonymous

60 defgrees

30. anonymous

the very bottom of the cheat sheet i sent has all the angles associated with the unit circle, and all the coordinates. look on the unit circle to see where the first coordinate is 0

31. anonymous

0,1

32. anonymous

90

33. anonymous

yay

34. anonymous

if you are working in degrees it is 90

35. anonymous

radians $\frac{\pi}{2}$

36. myininaya

is there another between 0 and 2pi, that cosine will be zero?

37. anonymous

but on my wksht theres noone of those answers.

38. anonymous

yes at 270

39. anonymous

whatever you prefer. so that takes care of $cos(\theta)=0$ for the moment. now we need $cos(\theta)=\frac{1}{2}$

40. anonymous

now that one is 60

41. anonymous

yes 90 works, and 270 also works

42. anonymous

or 300 degrees

43. anonymous

exactly

44. myininaya

or 3pi/2 now we can get cosine is 0 when we start at these angles and go around the circle again (2pi) so the answer for cos(t)=0 is t=2npi+pi/2 and t=2npi+3pi/2

45. anonymous

you are working in degrees so you have the answers.

46. myininaya

that whole paragraph is still talking about the first one k?

47. anonymous

theo nly answers available are pi over 6 0 pi/4 or pi/m3

48. anonymous

then they are missing $\frac{\pi}{2}$

49. anonymous

$\frac{\pi}{3}$ works, from the cheat sheet

50. anonymous

51. anonymous

$2 \cos^2 \theta- \cos \theta = 0$

52. anonymous

i know but i like using degrees better.

53. anonymous

so was pi/3 the answer? vecause cos was 1/2?

54. anonymous

yes

55. anonymous

i semi get it,thank you

56. anonymous

i guess it just asked 'which one of these is a solution"

57. anonymous

did you look at the cheat sheet? because it is a good one

58. anonymous

yeah which value satisfies the the equation

59. anonymous

yes, it had a lot of good material thank you

60. anonymous

could you helo me with my other math questions?

61. anonymous

2 cos_th^2 - cos_th = 0 cos_th (2 cos_th - 1) = 0 -> cos_th =0 or 1/2 -> th = theta = +pi/2 or - pi/2 or -pi/3 or +pi/3 *

62. anonymous

@LynFran