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anonymous

  • 5 years ago

which value of theta satisfies the equation 2cos^2theta - cos theta=0

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  1. anonymous
    • 5 years ago
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    \[cos^2(\theta)-cos(\theta)=0\] \[cos(\theta)(cos(\theta)-1)=0\] \[cos(\theta)=0\] or \[cos(\theta)=1\]

  2. anonymous
    • 5 years ago
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    oooooooops forgot the two!

  3. anonymous
    • 5 years ago
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    ignore my post, or just modify with the two.

  4. anonymous
    • 5 years ago
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    is there a formula to memorize forthis?

  5. anonymous
    • 5 years ago
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    \[cos(\theta)(2cos(\theta)-1)=0\] \[cos(\theta)=0\] \[cos(\theta)=\frac{1}{2}\]

  6. anonymous
    • 5 years ago
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    oh no the first part has nothing at all to do with trig

  7. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    like solving \[2x^2-x=0\] \[x(2x-1)=0\] \[x=0\] \[x=\frac{1}{2}\]

  9. anonymous
    • 5 years ago
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    oh no im confused

  10. anonymous
    • 5 years ago
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    the trig part comes now.

  11. myininaya
    • 5 years ago
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    oops that 11pi/6 is wrong its 5pi/3

  12. anonymous
    • 5 years ago
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    the first job is to solve for \[cos(\theta)\] the next step is to solve for \[\theta\]

  13. anonymous
    • 5 years ago
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    the first steps have nothing at all to do with trig. no formulas, no nothing.

  14. anonymous
    • 5 years ago
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    ok

  15. anonymous
    • 5 years ago
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    think of \[cos(\theta)\] as x and solve for x. you have \[2x^2-x=0\] and this is easy to solve

  16. anonymous
    • 5 years ago
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    we get \[x=0\] or \[x=\frac{1}{2}\] so now \[cos(\theta)=0\] or \[cos(\theta)=\frac{1}{2}\]

  17. anonymous
    • 5 years ago
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    the trig part comes now, seeing if we can find numbers for theta that will make this work.

  18. anonymous
    • 5 years ago
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    can you do the next step, that is, can you find theta?

  19. anonymous
    • 5 years ago
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    im not sure/

  20. anonymous
    • 5 years ago
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    isnt it 0?

  21. anonymous
    • 5 years ago
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    ok so first off we want to find where \[cos(\theta)=0\]

  22. anonymous
    • 5 years ago
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    http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

  23. myininaya
    • 5 years ago
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    just look between 0 and 2pi first to see where cosine is 0

  24. anonymous
    • 5 years ago
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    best cheat sheet. look on it to see where cosine is 0. it is not at 0, because \[cos(0)=1\]

  25. myininaya
    • 5 years ago
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    then after just add 2npi for n=0,1,2,3,... blah blah

  26. anonymous
    • 5 years ago
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    cos is the x correct?

  27. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    what happens at pi/2?

  29. anonymous
    • 5 years ago
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    60 defgrees

  30. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    0,1

  32. anonymous
    • 5 years ago
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    90

  33. anonymous
    • 5 years ago
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    yay

  34. anonymous
    • 5 years ago
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    if you are working in degrees it is 90

  35. anonymous
    • 5 years ago
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    radians \[\frac{\pi}{2}\]

  36. myininaya
    • 5 years ago
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    is there another between 0 and 2pi, that cosine will be zero?

  37. anonymous
    • 5 years ago
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    but on my wksht theres noone of those answers.

  38. anonymous
    • 5 years ago
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    yes at 270

  39. anonymous
    • 5 years ago
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    whatever you prefer. so that takes care of \[cos(\theta)=0\] for the moment. now we need \[cos(\theta)=\frac{1}{2}\]

  40. anonymous
    • 5 years ago
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    now that one is 60

  41. anonymous
    • 5 years ago
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    yes 90 works, and 270 also works

  42. anonymous
    • 5 years ago
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    or 300 degrees

  43. anonymous
    • 5 years ago
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    exactly

  44. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    you are working in degrees so you have the answers.

  46. myininaya
    • 5 years ago
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    that whole paragraph is still talking about the first one k?

  47. anonymous
    • 5 years ago
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    theo nly answers available are pi over 6 0 pi/4 or pi/m3

  48. anonymous
    • 5 years ago
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    then they are missing \[\frac{\pi}{2}\]

  49. anonymous
    • 5 years ago
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    \[\frac{\pi}{3}\] works, from the cheat sheet

  50. anonymous
    • 5 years ago
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    if those are your possible answers you are working in radians not degrees.

  51. anonymous
    • 5 years ago
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    \[2 \cos^2 \theta- \cos \theta = 0\]

  52. anonymous
    • 5 years ago
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    i know but i like using degrees better.

  53. anonymous
    • 5 years ago
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    so was pi/3 the answer? vecause cos was 1/2?

  54. anonymous
    • 5 years ago
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    yes

  55. anonymous
    • 5 years ago
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    i semi get it,thank you

  56. anonymous
    • 5 years ago
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    i guess it just asked 'which one of these is a solution"

  57. anonymous
    • 5 years ago
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    did you look at the cheat sheet? because it is a good one

  58. anonymous
    • 5 years ago
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    yeah which value satisfies the the equation

  59. anonymous
    • 5 years ago
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    yes, it had a lot of good material thank you

  60. anonymous
    • 5 years ago
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    could you helo me with my other math questions?

  61. L
    • 5 years ago
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    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
    • one year ago
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    @LynFran

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