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kenneyfamily

  • 3 years ago

Solve 4sin^2x + (4√2)cos x - 6 = 0 for 0 < x < 2π a. 3π/4, 7π/4 b. π/4, 7π/4 c. π/4, 5π/4 d. 3π/4, 5π/4

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  1. godorovg
    • 3 years ago
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    you there

  2. kenneyfamily
    • 3 years ago
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    yes

  3. godorovg
    • 3 years ago
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    okay tell me what you know first?

  4. kenneyfamily
    • 3 years ago
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    i don't even know way to start. trig functions really confuse me

  5. godorovg
    • 3 years ago
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    okay let me ask you do you know the unit circle?

  6. kenneyfamily
    • 3 years ago
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    yes!

  7. godorovg
    • 3 years ago
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    recall the untit circle and where sin is and so forth

  8. kenneyfamily
    • 3 years ago
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    its the y coordinate

  9. godorovg
    • 3 years ago
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    yes

  10. godorovg
    • 3 years ago
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    now let's look at the problem e 4sin^2x + (4√2)cos x - 6 = 0 for 0 < x < 2π tell what the 0>x>2 pi means?

  11. kenneyfamily
    • 3 years ago
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    im not sure...

  12. godorovg
    • 3 years ago
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    that is the limit or where this anser would fall on the circle

  13. godorovg
    • 3 years ago
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    okay now to slove the problem 4sin^2x what can be done here?

  14. kenneyfamily
    • 3 years ago
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    can you simplify it?

  15. godorovg
    • 3 years ago
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    yes

  16. kenneyfamily
    • 3 years ago
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    to what? sorry

  17. godorovg
    • 3 years ago
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    one sec

  18. zepdrix
    • 3 years ago
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    @godorovg If you're stuck, let me know :3 I can provide some assistance. I don't wanna step on your toes if you've got this though :)

  19. godorovg
    • 3 years ago
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    yeah I could use some help here I am tying to think how to do this in a simple way can you retrice me?

  20. zepdrix
    • 3 years ago
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    \[\large \sin^2 x + \cos^2 x = 1\]\[\large \sin^2 x = 1-\cos^2 x\] Using this second identity, we can write the entire equation in terms of cosines, and then treat it as a quadratic equation from there. Using the quadratic equation if necessary :D

  21. zepdrix
    • 3 years ago
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    Quadratic formula* blah i always mix those two up.. i forget which is which :3

  22. godorovg
    • 3 years ago
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    if you look at answers I think it is way first way

  23. zepdrix
    • 3 years ago
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    Oh an alternative, possibly easier method, since this is multiple choice, might be to just plug in your values given in A and see if it equals 0 or not. Just process of elimination :)

  24. zepdrix
    • 3 years ago
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    Although on a test that might not work so well :3 so you might wanna learn the method hehe

  25. godorovg
    • 3 years ago
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    this isn't me asking this question I was trying to help only

  26. zepdrix
    • 3 years ago
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    I know, i was talking to kenney ^^ my bad

  27. godorovg
    • 3 years ago
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    okay sorry I thought you were speaking to me

  28. zepdrix
    • 3 years ago
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    Using this identity:\[\large \sin^2 x = (1-\cos^2 x)\] Our equation becomes:\[\large 4\sin^2 x +4\sqrt2 \cos x - 6 = 0\]\[\large 4(1-\cos^2 x) +4\sqrt2 \cos x - 6 = 0\]

  29. zepdrix
    • 3 years ago
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    Understand what we did for the first step kenney? :O it's a little tricky, we're taking advantage of a trig identity.

  30. zepdrix
    • 3 years ago
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    kenney too quiet -_- hehe

  31. kenneyfamily
    • 3 years ago
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    sorry haha just trying to read all this and take it in

  32. kenneyfamily
    • 3 years ago
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    but yes i understand

  33. godorovg
    • 3 years ago
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    if we go too fast for you tell us okay

  34. godorovg
    • 3 years ago
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    answer choices are: a. 3π/4, 7π/4 b. π/4, 7π/4 c. π/4, 5π/4 d. 3π/4, 5π/4

  35. godorovg
    • 3 years ago
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    if take the unit circle and find these than you can slove this that way

  36. kenneyfamily
    • 3 years ago
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    how will locating on the unit circle help?

  37. godorovg
    • 3 years ago
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    if you know sin and cos and sec it's in you can use these values and use them

  38. zepdrix
    • 3 years ago
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    \[\large 4(1-\cos^2 x) +4\sqrt2 \cos x - 6 = 0\] Distributing the 4 gives us:\[\large 4-4\cos^2 x +4\sqrt2 \cos x - 6 = 0\]\[\large -4\cos^2 x +4\sqrt2 \cos x - 2 = 0\] Dividing both sides by -1 gives us:\[\large 4\cos^2 x-4\sqrt2 \cos x +2 = 0\] From here, we can think of cosx as something simple like U and it might be easier for you to solve it from there, using the quadratic formula. \[\large 4u^2-4\sqrt2u+2=0\] \[\large u=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

  39. zepdrix
    • 3 years ago
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    These are some of the steps you would take to solve this. It's a little tricky, it touches on trig AND some algebra stuff.. take a look at steps and maybe ask questions if you're confused by anything :D

  40. godorovg
    • 3 years ago
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    wait I am trying but failing to see something here??

  41. zepdrix
    • 3 years ago
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    I'm not exactly sure where you are with this material, so I figure I'd at least show you how to get the solution and maybe you can work though it at your own pace.

  42. zepdrix
    • 3 years ago
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    what godo? :D

  43. kenneyfamily
    • 3 years ago
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    sorry i'm lost..how will the quadratic formula help?

  44. zepdrix
    • 3 years ago
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    maybe it was unnecessary to change to U, I hope that didn't confuse you. U=cosx in this case. So we're solving for cosx. The quadratic formula will end up giving us a special angle.

  45. zepdrix
    • 3 years ago
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    A value that corresponds to one of the special angles i mean*

  46. kenneyfamily
    • 3 years ago
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    ok

  47. zepdrix
    • 3 years ago
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    \[\large \cos x = \frac{ 4\sqrt2 \pm \sqrt{(4\sqrt2)^2-4(4)(2)} }{ 2(4) }\] \[\large \cos x = \frac{ 4\sqrt2 \pm \sqrt{32-32} }{ 8 }\]\[\large \cos x = \frac{\sqrt2}{2}\]

  48. zepdrix
    • 3 years ago
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    It's kinda tricky to get there, but see what we did? :O We got a nice number that refers back to one of our reference angles.

  49. godorovg
    • 3 years ago
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    zep answers are these a. 3π/4, 7π/4 b. π/4, 7π/4 c. π/4, 5π/4 d. 3π/4, 5π/4 help me to see this??

  50. zepdrix
    • 3 years ago
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    So cos x is a POSITIVE sqrt(2)/2 at ... pi/4.. andddd... anyone remember the other one? :D

  51. kenneyfamily
    • 3 years ago
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    7pi/4?

  52. zepdrix
    • 3 years ago
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    yay \:D/

  53. kenneyfamily
    • 3 years ago
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    thanks!!!! you guys are awesome!!! :D

  54. godorovg
    • 3 years ago
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    zep that was what I was doing with the unit circle.

  55. zepdrix
    • 3 years ago
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    If there is a simpler way to get through this problem, then I apologize :D But this is how I remember doing it. As mentioned before, you can always try plugging your options into your initial equation and see which ones give you 0 :O

  56. godorovg
    • 3 years ago
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    no I think thanks for the help zep I am sorry I guess I am still rusty llittle forgive me Kenndy

  57. kenneyfamily
    • 3 years ago
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    no worries..all help was appreciated!

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