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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
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

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godorovgBest ResponseYou've already chosen the best response.0
okay tell me what you know first?
 one year ago

kenneyfamilyBest ResponseYou've already chosen the best response.0
i don't even know way to start. trig functions really confuse me
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
okay let me ask you do you know the unit circle?
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
recall the untit circle and where sin is and so forth
 one year ago

kenneyfamilyBest ResponseYou've already chosen the best response.0
its the y coordinate
 one year ago

godorovgBest ResponseYou've already chosen the best response.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?
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
that is the limit or where this anser would fall on the circle
 one year ago

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

kenneyfamilyBest ResponseYou've already chosen the best response.0
can you simplify it?
 one year ago

zepdrixBest 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 :)
 one year ago

godorovgBest ResponseYou've already chosen the best response.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?
 one year ago

zepdrixBest 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
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Quadratic formula* blah i always mix those two up.. i forget which is which :3
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
if you look at answers I think it is way first way
 one year ago

zepdrixBest ResponseYou've already chosen the best response.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 :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Although on a test that might not work so well :3 so you might wanna learn the method hehe
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
this isn't me asking this question I was trying to help only
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
I know, i was talking to kenney ^^ my bad
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
okay sorry I thought you were speaking to me
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Using this identity:\[\large \sin^2 x = (1\cos^2 x)\] Our equation becomes:\[\large 4\sin^2 x +4\sqrt2 \cos x  6 = 0\]\[\large 4(1\cos^2 x) +4\sqrt2 \cos x  6 = 0\]
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.1
kenney too quiet _ hehe
 one year ago

kenneyfamilyBest ResponseYou've already chosen the best response.0
sorry haha just trying to read all this and take it in
 one year ago

kenneyfamilyBest ResponseYou've already chosen the best response.0
but yes i understand
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
if we go too fast for you tell us okay
 one year ago

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

godorovgBest ResponseYou've already chosen the best response.0
if take the unit circle and find these than you can slove this that way
 one year ago

kenneyfamilyBest ResponseYou've already chosen the best response.0
how will locating on the unit circle help?
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
if you know sin and cos and sec it's in you can use these values and use them
 one year ago

zepdrixBest 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. \[\large 4u^24\sqrt2u+2=0\] \[\large u=\frac{ b \pm \sqrt{b^24ac} }{ 2a }\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.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
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
wait I am trying but failing to see something here??
 one year ago

zepdrixBest ResponseYou've already chosen the best response.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.
 one year ago

kenneyfamilyBest ResponseYou've already chosen the best response.0
sorry i'm lost..how will the quadratic formula help?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
A value that corresponds to one of the special angles i mean*
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.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.
 one year ago

godorovgBest ResponseYou've already chosen the best response.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??
 one year ago

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

kenneyfamilyBest ResponseYou've already chosen the best response.0
thanks!!!! you guys are awesome!!! :D
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
zep that was what I was doing with the unit circle.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.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
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
no I think thanks for the help zep I am sorry I guess I am still rusty llittle forgive me Kenndy
 one year ago

kenneyfamilyBest ResponseYou've already chosen the best response.0
no worries..all help was appreciated!
 one year ago
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