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onegirl Group Title

2 cos^2 x - 3 cos x + 1 = 0 for 0 is less than or equal to x is less than 2pi

  • one year ago
  • one year ago

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  1. dpaInc Group Title
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    the first term with the cosine... is it: \(\large 2cos^2x \) or \(\large 2cos(2x) \) i'm thinking it's the first one?

    • one year ago
  2. onegirl Group Title
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    yea the first one

    • one year ago
  3. dpaInc Group Title
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    think of it this way.. let y=cosx so your equation becomes: \(\large 2y^2-3y+1=0 \) can you solve this quadratic?

    • one year ago
  4. onegirl Group Title
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    yea i can

    • one year ago
  5. dpaInc Group Title
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    what is/are the solutions ? y = ???

    • one year ago
  6. z3529080 Group Title
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    cos2x=2cos^2(x)-1

    • one year ago
  7. z3529080 Group Title
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    2(2cos^2(x)-1)-3cosx+1=0 -> 4cos^2(x)-3cosx-1=0 -> (4cosx+1)(cosx-1)=0

    • one year ago
  8. onegirl Group Title
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    x1 = -0.5 x2 = -1

    • one year ago
  9. z3529080 Group Title
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    so set 4cosx+1=0 or cos-1 =0

    • one year ago
  10. dpaInc Group Title
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    hmmm.. i got the same but POSITIVE values... i'm double checking...

    • one year ago
  11. dpaInc Group Title
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    yeah... they should be positive: \(\large 2y^2-3y+1=0 \) \(\large (2y-1)(y-1)=0 \) so 2y - 1= 0 gives y=1/2 y-1 = 0 gives y=1

    • one year ago
  12. dpaInc Group Title
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    ok so far?

    • one year ago
  13. onegirl Group Title
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    yea

    • one year ago
  14. z3529080 Group Title
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    you forgot to x2

    • one year ago
  15. z3529080 Group Title
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    it is 2cos2x

    • one year ago
  16. onegirl Group Title
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    it is 2 cos^2 x - 3

    • one year ago
  17. z3529080 Group Title
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    oh i thought it is cos2x

    • one year ago
  18. dpaInc Group Title
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    since y=cosx, we have these two equations: \( \large cosx=\frac{1}{2}\) and \(\large cosx=-1 \) can you solve these?

    • one year ago
  19. dpaInc Group Title
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    oops... that second one should be POSITIVE one...

    • one year ago
  20. dpaInc Group Title
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    do you use the unit circle?

    • one year ago
  21. onegirl Group Title
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    yes i do

    • one year ago
  22. joshi Group Title
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    this is probably a stupid question, but i'm a bit confused. you said let y = cosx so what happens to the first term 2cos2x? like what about the 2 infront of the x?

    • one year ago
  23. onegirl Group Title
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    @joshi its 2 cos^2 x -3 i made a mistake in typing it

    • one year ago
  24. dpaInc Group Title
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    ok... so look at what angle gives you a cosine of 1/2.... HINT: there are two of 'em from 0 to 2pi

    • one year ago
  25. joshi Group Title
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    ohhh ok that makes more sense haha

    • one year ago
  26. dpaInc Group Title
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    yeah... i got a clarification from the asker with my first question... :)

    • one year ago
  27. onegirl Group Title
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    60 degrees?

    • one year ago
  28. dpaInc Group Title
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    |dw:1350069794735:dw| yep... 60 degrees is one of 'em....

    • one year ago
  29. onegirl Group Title
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    or pi/3

    • one year ago
  30. dpaInc Group Title
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    you got the second angle?

    • one year ago
  31. onegirl Group Title
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    and 45 degrees?

    • one year ago
  32. dpaInc Group Title
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    nope... we're still solving cosx = 1/2 , right? x=60 degrees or pi/3 is one... the second solution is in the fourth quadrant...

    • one year ago
  33. dpaInc Group Title
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    |dw:1350070145499:dw|

    • one year ago
  34. dpaInc Group Title
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    or take a look at page 3: http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

    • one year ago
  35. onegirl Group Title
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    240 degrees?

    • one year ago
  36. dpaInc Group Title
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    240 degrees is in the third quadrant... this is the fourth quadrant: |dw:1350070347242:dw|

    • one year ago
  37. onegirl Group Title
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    ohh ok 300 degrees..

    • one year ago
  38. dpaInc Group Title
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    yes... :) cos(60 degrees) = 1/2 and cos(300 degrees) = 1/2 now solve the other equation: cos x = 1

    • one year ago
  39. dpaInc Group Title
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    and remember, \(\large 0\color {red}{\le} x<2\pi \)

    • one year ago
  40. dpaInc Group Title
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    there should be only one solution for this one....

    • one year ago
  41. dpaInc Group Title
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    using the unit circle, what angle gives you an x-coordinate of 1?

    • one year ago
  42. onegirl Group Title
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    360 degrees?

    • one year ago
  43. dpaInc Group Title
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    yes but 2pi is not included in the interval you're looking at.. 360 degrees = 2pi so what other angle gives you an x-coordinate of 1?

    • one year ago
  44. dpaInc Group Title
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    HINT: \(\huge \le \)

    • one year ago
  45. onegirl Group Title
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    ohhh yea it will 180 degrees

    • one year ago
  46. dpaInc Group Title
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    no.. cos(180 degrees) = -1 try again....

    • one year ago
  47. dpaInc Group Title
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    cos x = 1 where \( \large 0\color {red}{\le} x<2\pi\)

    • one year ago
  48. onegirl Group Title
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    3pi/2?

    • one year ago
  49. dpaInc Group Title
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    what about the left side of \(\large 0\color {red}{\le} x<2\pi \) ???? x = 0 ? cos 0 = ???

    • one year ago
  50. dpaInc Group Title
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    ok... let's put it this way... what's an angle that is coterminal with 360 degrees?

    • one year ago
  51. onegirl Group Title
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    positive 360 and negative -360

    • one year ago
  52. dpaInc Group Title
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    no.. the angle you want is 0 because cos(0 degrees) = 1 so you have 3 solutions: x=0, 60, 300 degrees but since the problem was stated in terms of radians, i'd convert those angles.

    • one year ago
  53. onegirl Group Title
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    so it will be x=1.823 radian

    • one year ago
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