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anonymous
 4 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
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay tell me what you know first?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't even know way to start. trig functions really confuse me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay let me ask you do you know the unit circle?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0recall the untit circle and where sin is and so forth

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now 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
 4 years ago
Best ResponseYou've already chosen the best response.0that is the limit or where this anser would fall on the circle

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay now to slove the problem 4sin^2x what can be done here?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.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 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah I could use some help here I am tying to think how to do this in a simple way can you retrice me?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.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

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Quadratic formula* blah i always mix those two up.. i forget which is which :3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you look at answers I think it is way first way

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Oh 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
 4 years ago
Best ResponseYou've already chosen the best response.1Although on a test that might not work so well :3 so you might wanna learn the method hehe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this isn't me asking this question I was trying to help only

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1I know, i was talking to kenney ^^ my bad

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay sorry I thought you were speaking to me

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Using 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
 4 years ago
Best ResponseYou've already chosen the best response.1Understand what we did for the first step kenney? :O it's a little tricky, we're taking advantage of a trig identity.

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1kenney too quiet _ hehe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry haha just trying to read all this and take it in

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if we go too fast for you tell us okay

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0answer choices are: a. 3π/4, 7π/4 b. π/4, 7π/4 c. π/4, 5π/4 d. 3π/4, 5π/4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if take the unit circle and find these than you can slove this that way

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how will locating on the unit circle help?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you know sin and cos and sec it's in you can use these values and use them

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1\[\large 4(1\cos^2 x) +4\sqrt2 \cos x  6 = 0\] Distributing the 4 gives us:\[\large 44\cos^2 x +4\sqrt2 \cos x  6 = 0\]\[\large 4\cos^2 x +4\sqrt2 \cos x  2 = 0\] Dividing both sides by 1 gives us:\[\large 4\cos^2 x4\sqrt2 \cos x +2 = 0\] 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. \[\large 4u^24\sqrt2u+2=0\] \[\large u=\frac{ b \pm \sqrt{b^24ac} }{ 2a }\]

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1These 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
 4 years ago
Best ResponseYou've already chosen the best response.0wait I am trying but failing to see something here??

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1I'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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry i'm lost..how will the quadratic formula help?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1maybe 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
 4 years ago
Best ResponseYou've already chosen the best response.1A value that corresponds to one of the special angles i mean*

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1\[\large \cos x = \frac{ 4\sqrt2 \pm \sqrt{(4\sqrt2)^24(4)(2)} }{ 2(4) }\] \[\large \cos x = \frac{ 4\sqrt2 \pm \sqrt{3232} }{ 8 }\]\[\large \cos x = \frac{\sqrt2}{2}\]

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1It'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
 4 years ago
Best ResponseYou've already chosen the best response.0zep 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
 4 years ago
Best ResponseYou've already chosen the best response.1So cos x is a POSITIVE sqrt(2)/2 at ... pi/4.. andddd... anyone remember the other one? :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks!!!! you guys are awesome!!! :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0zep that was what I was doing with the unit circle.

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1If 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
 4 years ago
Best ResponseYou've already chosen the best response.0no I think thanks for the help zep I am sorry I guess I am still rusty llittle forgive me Kenndy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no worries..all help was appreciated!
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