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anonymous

  • one year ago

Use a graphing calculator to solve the equation -3 cos t = 1 in the interval from

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  1. anonymous
    • one year ago
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    \[0\le \theta \le 2x\]

  2. anonymous
    • one year ago
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    Round to the nearest hundredth and show your work.

  3. freckles
    • one year ago
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    is x suppose to be pi?

  4. anonymous
    • one year ago
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    oh yeah sorry it is

  5. freckles
    • one year ago
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    \[\cos(t)=\frac{-1}{3} \text{ on } 0 \le t \le 2\pi \\ \text{ one solution can be found just by taking } \arccos( ) \text{ of both sides } \\ \text{ other one we can find \it by using that } \\ \text{ cosine is even and has period } 2 \pi \\ \cos(-t)=\cos(t) \\ \text{ so we have } \\ \cos(-t+2\pi)=\frac{-1}{3} \\ \text{ take } \arccos( ) \text{ of both sides and solve for } t \]

  6. anonymous
    • one year ago
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    @freckles can you keep going cuz i really dont know how to do this

  7. anonymous
    • one year ago
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    I fanned you and gave you a medal

  8. freckles
    • one year ago
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    well were you not able to the first solution? it is just taking arccos( ) of both sides

  9. freckles
    • one year ago
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    \[\cos(t)=\frac{-1}{3} \text{ we could say one solution is } t=\arccos(\frac{-1}{3})\]

  10. anonymous
    • one year ago
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    No im really confused, that's why i like when people go step by step

  11. freckles
    • one year ago
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    Oh I thought I gave you step by step

  12. freckles
    • one year ago
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    all I did there was take arccos( ) of both sides for that one solution

  13. freckles
    • one year ago
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    just as I said to do in the steps above

  14. anonymous
    • one year ago
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    you did, but is that the answer?

  15. freckles
    • one year ago
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    that is one solution i also said how to find the other

  16. freckles
    • one year ago
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    use the fact that cos is even and has period 2pi

  17. anonymous
    • one year ago
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    so the answer is -1/3?

  18. freckles
    • one year ago
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    how did you get that?

  19. anonymous
    • one year ago
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    No i'm asking you, from your step by step that's what is above that everything is equalling -1/3

  20. freckles
    • one year ago
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    the equation is cos(t)=-1/3 we aren't solving for cos(t) if we were then we would done and the answer is -1/3 we are solving for t

  21. freckles
    • one year ago
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    one solution is given by taking arccos( ) of both sides of the equation cos(t)=-1/3

  22. anonymous
    • one year ago
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    ok how do we do that?

  23. freckles
    • one year ago
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    one solution is given by t=arccos(-1/3)

  24. freckles
    • one year ago
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    and then the other solution is given by using the fact that cos is even and has period 2pi

  25. anonymous
    • one year ago
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    t=4.37255207±2πn

  26. freckles
    • one year ago
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    let me give you an easy example to follow: \[\cos(\theta)=\frac{-1}{2} , \text{ where } 0 \le \theta \le 2 \pi \\ \text{ we know we are suppose to get solutions } \theta=\frac{2 \pi }{3} ,\frac{4\pi}{3} \\ \text{ we see we can get these solutions by doing : } \\ \text{ Answer one given by just takinng } \arccos( ) \text{ of both sides } \\ \theta=\arccos(\frac{-1}{2}) \\ \\ \text{ now the other solution comes from using the fact that } \\ \text{ cosine is even function and has period } 2\pi \\ \cos(-\theta+2\pi)=\frac{-1}{2} \\ \text{ solve for } \theta \text{ by first taking } \arccos( ) \text{ of both sides } \\ -\theta+2\pi=\arccos(\frac{-1}{2}) \\ \text{ now subtract } 2\pi \text{ on both sides } \\ -\theta=\arccos(\frac{-1}{2})-2\pi \\ \text{ last step multiply -1 on both sides } \\ \theta=-\arccos(\frac{-1}{2})+2\pi \\ \\ \text{ so the solutions to } \cos(\theta)=\frac{-1}{2} \text{ on } 0 \le \theta \le 2\pi \text{ are } \\ \theta=\arccos(\frac{-1}{2}) \text{ or } \theta=-\arccos(\frac{-1}{2})+2\pi \\ \text{ we can simplify these solutions } \\ \text{ We know } \arccos(\frac{-1}{2})=\frac{2\pi}{3} \\ \\\ \text{ so we have } \theta=\frac{2\pi}{3} \text{ or } \theta=-\frac{2\pi}{3}+2\pi=\frac{4\pi}{3} \\ \text{ just as we got way above just from using the unit circle }\]

  27. freckles
    • one year ago
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    |dw:1437253274947:dw|

  28. freckles
    • one year ago
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    anyways you can use the exact same process I used above

  29. freckles
    • one year ago
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    just take arccos() of both sides to get one solution

  30. freckles
    • one year ago
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    then rewrite the equation as cos(-t+2pi)=whatever where whatever is between -1 and 1 (inclusive) and take arccos( ) of both sides and then solve that resulting equation for t

  31. freckles
    • one year ago
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    \[\cos(t)=\frac{-1}{3} \\ \text{ one solution is } t=\arccos(\frac{-1}{3} ) \\ \text{ the other solution comes from solving } \cos(-t+2\pi)=\frac{-1}{3}\]

  32. freckles
    • one year ago
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    if I take arccos( ) of both sides of your second equation -t+2pi=arccos(-1/3) can you try to solve this for t?

  33. freckles
    • one year ago
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    it is just two steps

  34. anonymous
    • one year ago
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    Is the answer theta 4/2pi?

  35. freckles
    • one year ago
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    do you know how to solve -x+4=6 for x?

  36. anonymous
    • one year ago
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    yes

  37. freckles
    • one year ago
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    like you would subtract 4 on both sides -x=6-4 -x=2 and last step is multiply -1 on both sides x=-2 like this is the same type of equation here: \[-t+2\pi=\arccos(\frac{-1}{3} ) \\ \text{ subtract } 2\pi \text{ on both sides } \\ -t =\arccos(\frac{-1}{3})-2\pi \\ \text{ multiply -1 on both sides } \\ t=-\arccos(\frac{-1}{3})+2\pi\]

  38. freckles
    • one year ago
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    you have both solutions you can use your calculator to approximate them if you want

  39. freckles
    • one year ago
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    I think the instructions above do say to round to nearest hundredth so you will have to break out the calculator

  40. anonymous
    • one year ago
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    Will you help me to the final answer?

  41. freckles
    • one year ago
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    I gave you both exact solutions... \[t=\arccos(\frac{-1}{3}) \\ t=-\arccos(\frac{-1}{3})+2\pi\] ?

  42. freckles
    • one year ago
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    oh are you saying you don't know how to plug it in the calculator?

  43. freckles
    • one year ago
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    there should be arcos typed on your calculator right above the cos button

  44. freckles
    • one year ago
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    arccos*

  45. freckles
    • one year ago
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    are it could be written as cos^(-1)

  46. freckles
    • one year ago
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    do you see it?

  47. freckles
    • one year ago
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    to access it on some calculators you have to press (shift or 2nd) can't remember which one

  48. anonymous
    • one year ago
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    No i cant get a good answer it keeps saying error

  49. anonymous
    • one year ago
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    t=8.19381854

  50. freckles
    • one year ago
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    what calculator do you use i can try to look it up and tell you what buttons to push

  51. anonymous
    • one year ago
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    t=8.19381854

  52. anonymous
    • one year ago
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    is that right?

  53. freckles
    • one year ago
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    oh i was waiting on you to tell me the calculator type but no

  54. anonymous
    • one year ago
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    It's on a website cuz i dont have a calculator

  55. freckles
    • one year ago
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    arccos(-1/3)=1.91 by using a calculator so we already have that as a solution now replace the arccos(-1/3) in the other solution with 1.91 \[-1.91+2\pi\] you can do 3.14 as an approximation for pi 3.14*2=6.28 so what is -1.91+6.28

  56. anonymous
    • one year ago
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    4.37

  57. freckles
    • one year ago
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    yes so remember you have two answers

  58. anonymous
    • one year ago
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    So 4.37 is one, what's the other?

  59. freckles
    • one year ago
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    :(

  60. anonymous
    • one year ago
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    what?

  61. freckles
    • one year ago
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    I just feel like you haven't listened to me \[t=\arccos(\frac{-1}{3}) \approx 1.91 \\ t=-\arccos(\frac{-1}{3})+2\pi \approx -1.91+2(3.14) =4.37\]

  62. anonymous
    • one year ago
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    No I have, im just learning. Im sorry

  63. freckles
    • one year ago
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    ok well I hope you understood some of what I said

  64. freckles
    • one year ago
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    I have to leave now any questions before I go

  65. anonymous
    • one year ago
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    Im gonna overview it right now, thank you very much for your help i appreciate it!

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