anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
you there
anonymous
  • anonymous
yes
anonymous
  • anonymous
okay tell me what you know first?

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anonymous
  • anonymous
i don't even know way to start. trig functions really confuse me
anonymous
  • anonymous
okay let me ask you do you know the unit circle?
anonymous
  • anonymous
yes!
anonymous
  • anonymous
recall the untit circle and where sin is and so forth
anonymous
  • anonymous
its the y coordinate
anonymous
  • anonymous
yes
anonymous
  • anonymous
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?
anonymous
  • anonymous
im not sure...
anonymous
  • anonymous
that is the limit or where this anser would fall on the circle
anonymous
  • anonymous
okay now to slove the problem 4sin^2x what can be done here?
anonymous
  • anonymous
can you simplify it?
anonymous
  • anonymous
yes
anonymous
  • anonymous
to what? sorry
anonymous
  • anonymous
one sec
zepdrix
  • 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 :)
anonymous
  • anonymous
yeah I could use some help here I am tying to think how to do this in a simple way can you retrice me?
zepdrix
  • 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
zepdrix
  • zepdrix
Quadratic formula* blah i always mix those two up.. i forget which is which :3
anonymous
  • anonymous
if you look at answers I think it is way first way
zepdrix
  • 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 :)
zepdrix
  • zepdrix
Although on a test that might not work so well :3 so you might wanna learn the method hehe
anonymous
  • anonymous
this isn't me asking this question I was trying to help only
zepdrix
  • zepdrix
I know, i was talking to kenney ^^ my bad
anonymous
  • anonymous
okay sorry I thought you were speaking to me
zepdrix
  • zepdrix
Using this identity:\[\large \sin^2 x = (1-\cos^2 x)\] Our equation becomes:\[\large 4\sin^2 x +4\sqrt2 \cos x - 6 = 0\]\[\large 4(1-\cos^2 x) +4\sqrt2 \cos x - 6 = 0\]
zepdrix
  • zepdrix
Understand what we did for the first step kenney? :O it's a little tricky, we're taking advantage of a trig identity.
zepdrix
  • zepdrix
kenney too quiet -_- hehe
anonymous
  • anonymous
sorry haha just trying to read all this and take it in
anonymous
  • anonymous
but yes i understand
anonymous
  • anonymous
if we go too fast for you tell us okay
anonymous
  • anonymous
answer choices are: a. 3π/4, 7π/4 b. π/4, 7π/4 c. π/4, 5π/4 d. 3π/4, 5π/4
anonymous
  • anonymous
if take the unit circle and find these than you can slove this that way
anonymous
  • anonymous
how will locating on the unit circle help?
anonymous
  • anonymous
if you know sin and cos and sec it's in you can use these values and use them
zepdrix
  • 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. \[\large 4u^2-4\sqrt2u+2=0\] \[\large u=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
zepdrix
  • 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
anonymous
  • anonymous
wait I am trying but failing to see something here??
zepdrix
  • 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.
zepdrix
  • zepdrix
what godo? :D
anonymous
  • anonymous
sorry i'm lost..how will the quadratic formula help?
zepdrix
  • 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.
zepdrix
  • zepdrix
A value that corresponds to one of the special angles i mean*
anonymous
  • anonymous
ok
zepdrix
  • zepdrix
\[\large \cos x = \frac{ 4\sqrt2 \pm \sqrt{(4\sqrt2)^2-4(4)(2)} }{ 2(4) }\] \[\large \cos x = \frac{ 4\sqrt2 \pm \sqrt{32-32} }{ 8 }\]\[\large \cos x = \frac{\sqrt2}{2}\]
zepdrix
  • 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.
anonymous
  • anonymous
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??
zepdrix
  • zepdrix
So cos x is a POSITIVE sqrt(2)/2 at ... pi/4.. andddd... anyone remember the other one? :D
anonymous
  • anonymous
7pi/4?
zepdrix
  • zepdrix
yay \:D/
anonymous
  • anonymous
thanks!!!! you guys are awesome!!! :D
anonymous
  • anonymous
zep that was what I was doing with the unit circle.
zepdrix
  • 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
anonymous
  • anonymous
no I think thanks for the help zep I am sorry I guess I am still rusty llittle forgive me Kenndy
anonymous
  • anonymous
no worries..all help was appreciated!

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