Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

kenneyfamily

  • 2 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

  • This Question is Closed
  1. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you there

  2. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  3. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay tell me what you know first?

  4. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i don't even know way to start. trig functions really confuse me

  5. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay let me ask you do you know the unit circle?

  6. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes!

  7. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    recall the untit circle and where sin is and so forth

  8. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    its the y coordinate

  9. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  10. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im not sure...

  12. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is the limit or where this anser would fall on the circle

  13. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay now to slove the problem 4sin^2x what can be done here?

  14. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you simplify it?

  15. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  16. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    to what? sorry

  17. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    one sec

  18. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large \sin^2 x + \cos^2 x = 1\]\[\large \sin^2 x = 1-\cos^2 x\] 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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Quadratic formula* blah i always mix those two up.. i forget which is which :3

  22. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if you look at answers I think it is way first way

  23. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Although on a test that might not work so well :3 so you might wanna learn the method hehe

  25. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this isn't me asking this question I was trying to help only

  26. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I know, i was talking to kenney ^^ my bad

  27. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay sorry I thought you were speaking to me

  28. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    kenney too quiet -_- hehe

  31. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry haha just trying to read all this and take it in

  32. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but yes i understand

  33. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if we go too fast for you tell us okay

  34. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if take the unit circle and find these than you can slove this that way

  36. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how will locating on the unit circle help?

  37. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if you know sin and cos and sec it's in you can use these values and use them

  38. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large 4(1-\cos^2 x) +4\sqrt2 \cos x - 6 = 0\] 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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait I am trying but failing to see something here??

  41. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what godo? :D

  43. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry i'm lost..how will the quadratic formula help?

  44. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    A value that corresponds to one of the special angles i mean*

  46. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  47. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large \cos x = \frac{ 4\sqrt2 \pm \sqrt{(4\sqrt2)^2-4(4)(2)} }{ 2(4) }\] \[\large \cos x = \frac{ 4\sqrt2 \pm \sqrt{32-32} }{ 8 }\]\[\large \cos x = \frac{\sqrt2}{2}\]

  48. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  51. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    7pi/4?

  52. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yay \:D/

  53. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks!!!! you guys are awesome!!! :D

  54. godorovg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    zep that was what I was doing with the unit circle.

  55. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no I think thanks for the help zep I am sorry I guess I am still rusty llittle forgive me Kenndy

  57. kenneyfamily
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no worries..all help was appreciated!

  58. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.