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onegirl

  • 3 years ago

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

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  1. dpaInc
    • 3 years ago
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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?

  2. onegirl
    • 3 years ago
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    yea the first one

  3. dpaInc
    • 3 years ago
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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?

  4. onegirl
    • 3 years ago
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    yea i can

  5. dpaInc
    • 3 years ago
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    what is/are the solutions ? y = ???

  6. z3529080
    • 3 years ago
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    cos2x=2cos^2(x)-1

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

  8. onegirl
    • 3 years ago
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    x1 = -0.5 x2 = -1

  9. z3529080
    • 3 years ago
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    so set 4cosx+1=0 or cos-1 =0

  10. dpaInc
    • 3 years ago
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    hmmm.. i got the same but POSITIVE values... i'm double checking...

  11. dpaInc
    • 3 years ago
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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

  12. dpaInc
    • 3 years ago
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    ok so far?

  13. onegirl
    • 3 years ago
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    yea

  14. z3529080
    • 3 years ago
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    you forgot to x2

  15. z3529080
    • 3 years ago
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    it is 2cos2x

  16. onegirl
    • 3 years ago
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    it is 2 cos^2 x - 3

  17. z3529080
    • 3 years ago
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    oh i thought it is cos2x

  18. dpaInc
    • 3 years ago
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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?

  19. dpaInc
    • 3 years ago
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    oops... that second one should be POSITIVE one...

  20. dpaInc
    • 3 years ago
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    do you use the unit circle?

  21. onegirl
    • 3 years ago
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    yes i do

  22. joshi
    • 3 years ago
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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?

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

  24. dpaInc
    • 3 years ago
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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

  25. joshi
    • 3 years ago
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    ohhh ok that makes more sense haha

  26. dpaInc
    • 3 years ago
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    yeah... i got a clarification from the asker with my first question... :)

  27. onegirl
    • 3 years ago
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    60 degrees?

  28. dpaInc
    • 3 years ago
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    |dw:1350069794735:dw| yep... 60 degrees is one of 'em....

  29. onegirl
    • 3 years ago
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    or pi/3

  30. dpaInc
    • 3 years ago
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    you got the second angle?

  31. onegirl
    • 3 years ago
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    and 45 degrees?

  32. dpaInc
    • 3 years ago
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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...

  33. dpaInc
    • 3 years ago
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    |dw:1350070145499:dw|

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

  35. onegirl
    • 3 years ago
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    240 degrees?

  36. dpaInc
    • 3 years ago
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    240 degrees is in the third quadrant... this is the fourth quadrant: |dw:1350070347242:dw|

  37. onegirl
    • 3 years ago
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    ohh ok 300 degrees..

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

  39. dpaInc
    • 3 years ago
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    and remember, \(\large 0\color {red}{\le} x<2\pi \)

  40. dpaInc
    • 3 years ago
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    there should be only one solution for this one....

  41. dpaInc
    • 3 years ago
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    using the unit circle, what angle gives you an x-coordinate of 1?

  42. onegirl
    • 3 years ago
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    360 degrees?

  43. dpaInc
    • 3 years ago
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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?

  44. dpaInc
    • 3 years ago
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    HINT: \(\huge \le \)

  45. onegirl
    • 3 years ago
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    ohhh yea it will 180 degrees

  46. dpaInc
    • 3 years ago
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    no.. cos(180 degrees) = -1 try again....

  47. dpaInc
    • 3 years ago
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    cos x = 1 where \( \large 0\color {red}{\le} x<2\pi\)

  48. onegirl
    • 3 years ago
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    3pi/2?

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

  50. dpaInc
    • 3 years ago
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    ok... let's put it this way... what's an angle that is coterminal with 360 degrees?

  51. onegirl
    • 3 years ago
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    positive 360 and negative -360

  52. dpaInc
    • 3 years ago
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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.

  53. onegirl
    • 3 years ago
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    so it will be x=1.823 radian

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