## 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

1. godorovg

you there

2. kenneyfamily

yes

3. godorovg

okay tell me what you know first?

4. kenneyfamily

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

5. godorovg

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

6. kenneyfamily

yes!

7. godorovg

recall the untit circle and where sin is and so forth

8. kenneyfamily

its the y coordinate

9. godorovg

yes

10. godorovg

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

im not sure...

12. godorovg

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

13. godorovg

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

14. kenneyfamily

can you simplify it?

15. godorovg

yes

16. kenneyfamily

to what? sorry

17. godorovg

one sec

18. zepdrix

@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

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

$\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

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

22. godorovg

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

23. zepdrix

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

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

25. godorovg

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

26. zepdrix

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

27. godorovg

okay sorry I thought you were speaking to me

28. zepdrix

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

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

kenney too quiet -_- hehe

31. kenneyfamily

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

32. kenneyfamily

but yes i understand

33. godorovg

if we go too fast for you tell us okay

34. godorovg

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

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

36. kenneyfamily

how will locating on the unit circle help?

37. godorovg

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

38. zepdrix

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

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

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

41. zepdrix

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

what godo? :D

43. kenneyfamily

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

44. zepdrix

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

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

46. kenneyfamily

ok

47. zepdrix

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

48. zepdrix

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

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

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

51. kenneyfamily

7pi/4?

52. zepdrix

yay \:D/

53. kenneyfamily

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

54. godorovg

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

55. zepdrix

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

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

57. kenneyfamily

no worries..all help was appreciated!