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anonymous

  • one year ago

What values for,(posted below) satisfy the equation?

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  1. anonymous
    • one year ago
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    This fills in the blank

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  2. anonymous
    • one year ago
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    This follows the question

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  3. anonymous
    • one year ago
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    @UsukiDoll

  4. UsukiDoll
    • one year ago
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    I don't understand what the question needs besides from 0< x< 2pi means from 0 to 360 degrees which is the full circle

  5. UsukiDoll
    • one year ago
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    there is a trig identity \[\tan^2(\theta) = 1-\sec^2(\theta) \]

  6. anonymous
    • one year ago
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    What values for, |dw:1436466383880:dw| satisfy the equation?

  7. UsukiDoll
    • one year ago
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    maybe we can replace that tan with what I've wrote... ??? dah I've never seen this formatted question before x.x |dw:1436502474870:dw|

  8. anonymous
    • one year ago
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    The rest of the equation is in the second attachment at the top

  9. anonymous
    • one year ago
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    |dw:1436466531453:dw|

  10. UsukiDoll
    • one year ago
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    this is what I'm thinking but I'm not 100% sure ... use the identity on the left hand side \[\tan^2(\theta) = 1-\sec^2(\theta) \]

  11. UsukiDoll
    • one year ago
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    and then add \[\frac{3}{2}\sec(\theta) \] to both sides so we end up with some sort of quadratic equation (well similar )

  12. Oleg3321
    • one year ago
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    ehhh im not so sure how to do this one either.

  13. anonymous
    • one year ago
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    |dw:1436466790578:dw|

  14. anonymous
    • one year ago
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    Those are my answer options

  15. misty1212
    • one year ago
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    HI do what was suggested above write it as a quadratic equation in terms of secant then solve

  16. anonymous
    • one year ago
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    I don't know how to do these equations at all

  17. misty1212
    • one year ago
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    lol then cheat!

  18. anonymous
    • one year ago
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    haha well i was thinking you guys could give me the right answer cuz i cant find it

  19. misty1212
    • one year ago
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    i can show you how to find it if you like it will take a bit of time http://www.wolframalpha.com/input/?i=tan%5E2%28x%29%3D-3%2F2sec%28x%29

  20. UsukiDoll
    • one year ago
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    yes @misty1212 we write it as a quadratic equation in terms of secant and then solve

  21. anonymous
    • one year ago
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    @misty1212 anything can help, i just need the right answer

  22. UsukiDoll
    • one year ago
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    I know most of the process... but towards the end . I might crash

  23. UsukiDoll
    • one year ago
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    I think it has something to do with adding 2pi or subtracting 2pi ...hence more than one answer

  24. UsukiDoll
    • one year ago
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    |dw:1436503293700:dw|

  25. UsukiDoll
    • one year ago
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    |dw:1436503337897:dw|

  26. UsukiDoll
    • one year ago
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    |dw:1436503367107:dw| now multiply the entire equation by -2 to get rid of the fraction and the negative

  27. anonymous
    • one year ago
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    idk, i dont have a calculator at the moment sadly lol

  28. UsukiDoll
    • one year ago
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    |dw:1436503536319:dw|

  29. UsukiDoll
    • one year ago
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    |dw:1436503556177:dw|

  30. UsukiDoll
    • one year ago
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    now let all secants be x for now because that quadratic equation looks mean

  31. UsukiDoll
    • one year ago
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    |dw:1436503613776:dw|

  32. UsukiDoll
    • one year ago
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    oh yes we can factor this woo!

  33. anonymous
    • one year ago
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    ok im learning lol

  34. UsukiDoll
    • one year ago
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    |dw:1436503685977:dw|

  35. UsukiDoll
    • one year ago
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    I'm using b^2-4ac so I would know what that part is already for that quadratic formula

  36. UsukiDoll
    • one year ago
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    |dw:1436503730219:dw|

  37. UsukiDoll
    • one year ago
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    |dw:1436503768417:dw|

  38. UsukiDoll
    • one year ago
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    |dw:1436503804424:dw|

  39. anonymous
    • one year ago
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    So how do we find one of my answers from A-D? lol

  40. UsukiDoll
    • one year ago
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    \[(\sec(\theta)-2)(\sec(\theta)+\frac{1}{2})=0\] something is screwed up... I am expanding and the last term isn't working out

  41. anonymous
    • one year ago
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    @satellite73 sos

  42. UsukiDoll
    • one year ago
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    nice I found the problem -_-! only one of them is right

  43. UsukiDoll
    • one year ago
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    but strange taking the quadratic equation approach always works...

  44. anonymous
    • one year ago
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    Can you explain, or do you just wanna give the answer?

  45. UsukiDoll
    • one year ago
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    -__- *facepalm* I made that error by not getting rid of that fraction .. argh.. ok I thought we were done for.

  46. UsukiDoll
    • one year ago
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    \[(\sec(\theta)-2)(2\sec(\theta)+1)=0\]

  47. UsukiDoll
    • one year ago
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    so now we have to solve this separately . . . which I swear we have already done it when I found my roots.

  48. anonymous
    • one year ago
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    Alrightyyy

  49. UsukiDoll
    • one year ago
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    \[\sec(\theta)-2=0\] adding 2 to both sides \[\sec(\theta) =2 \] \[(2\sec(\theta)+1)=0 \] that's \[\sec(\theta) = \frac{-1}{2} \]

  50. UsukiDoll
    • one year ago
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    turns out that we already did this step and not even know it ^ because I've used the quadratic equation to get its roots.

  51. UsukiDoll
    • one year ago
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    so secant = 1/cosine this would be a bit easier to work with

  52. anonymous
    • one year ago
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    ok whats next

  53. UsukiDoll
    • one year ago
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    so my cosine has to be \[\frac{1}{2} \] for \[\sec(\theta) =2 \] and to check secant of 1/2 = 1/(1/2) = 2

  54. UsukiDoll
    • one year ago
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    |dw:1436504658208:dw| based on that triangle cosine 1/2 is 60 degrees otherwise known as \[\frac{\pi}{3} \] in radians

  55. anonymous
    • one year ago
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    So is the answer D?

  56. UsukiDoll
    • one year ago
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    yeah, but I want to get the other answer from that choice.

  57. UsukiDoll
    • one year ago
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    we got our pi/3. let's see if I can grab 5pi/3

  58. UsukiDoll
    • one year ago
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    which is 300 degrees.

  59. UsukiDoll
    • one year ago
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    OH MAN! I am making this hard on myself again ROFL! We just have to do this \[2\pi-\frac{\pi}{3} \] because according to our restriction we are in the unit circle 0< theta < 360 means from 0 to 360.. since we need 300 degrees we are in the fourth quadrant.

  60. UsukiDoll
    • one year ago
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    \[\frac{3}{3} \times 2\pi-\frac{\pi}{3} = \frac{6 \pi-\pi}{3} \]

  61. UsukiDoll
    • one year ago
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    \[\frac{5\pi}{3} \]

  62. UsukiDoll
    • one year ago
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    there. XDDDDDDDDDDDDDDDDDDDDDDD

  63. UsukiDoll
    • one year ago
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    I can't believe I just did that ... it's been 6 years....

  64. anonymous
    • one year ago
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    yay you did it lol

  65. UsukiDoll
    • one year ago
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    the wording on the question threw me off... because in my textbooks it was a bit simpler so I had to think it through for a bit.

  66. anonymous
    • one year ago
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    yeah its tough sometimes, but at least you guys are great help

  67. UsukiDoll
    • one year ago
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    thanks :D

  68. anonymous
    • one year ago
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    you're welcome. i gave you a medal

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