## kenneyfamily Group Title 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 one year ago one year ago

1. godorovg Group Title

you there

2. kenneyfamily Group Title

yes

3. godorovg Group Title

okay tell me what you know first?

4. kenneyfamily Group Title

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

5. godorovg Group Title

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

6. kenneyfamily Group Title

yes!

7. godorovg Group Title

recall the untit circle and where sin is and so forth

8. kenneyfamily Group Title

its the y coordinate

9. godorovg Group Title

yes

10. godorovg Group Title

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 Group Title

im not sure...

12. godorovg Group Title

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

13. godorovg Group Title

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

14. kenneyfamily Group Title

can you simplify it?

15. godorovg Group Title

yes

16. kenneyfamily Group Title

to what? sorry

17. godorovg Group Title

one sec

18. zepdrix Group Title

@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 Group Title

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 Group Title

$\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 Group Title

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

22. godorovg Group Title

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

23. zepdrix Group Title

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 Group Title

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

25. godorovg Group Title

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

26. zepdrix Group Title

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

27. godorovg Group Title

okay sorry I thought you were speaking to me

28. zepdrix Group Title

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 Group Title

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 Group Title

kenney too quiet -_- hehe

31. kenneyfamily Group Title

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

32. kenneyfamily Group Title

but yes i understand

33. godorovg Group Title

if we go too fast for you tell us okay

34. godorovg Group Title

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 Group Title

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

36. kenneyfamily Group Title

how will locating on the unit circle help?

37. godorovg Group Title

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

38. zepdrix Group Title

$\large 4(1-\cos^2 x) +4\sqrt2 \cos x - 6 = 0$ 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 Group Title

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 Group Title

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

41. zepdrix Group Title

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 Group Title

what godo? :D

43. kenneyfamily Group Title

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

44. zepdrix Group Title

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 Group Title

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

46. kenneyfamily Group Title

ok

47. zepdrix Group Title

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

48. zepdrix Group Title

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 Group Title

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 Group Title

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

51. kenneyfamily Group Title

7pi/4?

52. zepdrix Group Title

yay \:D/

53. kenneyfamily Group Title

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

54. godorovg Group Title

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

55. zepdrix Group Title

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 Group Title

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

57. kenneyfamily Group Title

no worries..all help was appreciated!