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

- schrodinger

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- anonymous

you there

- anonymous

yes

- anonymous

okay tell me what you know first?

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## More answers

- anonymous

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

- anonymous

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

- anonymous

yes!

- anonymous

recall the untit circle and where sin is and so forth

- anonymous

its the y coordinate

- anonymous

yes

- 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

im not sure...

- anonymous

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

- anonymous

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

- anonymous

can you simplify it?

- anonymous

yes

- anonymous

to what? sorry

- anonymous

one sec

- 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

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

\[\large \sin^2 x + \cos^2 x = 1\]\[\large \sin^2 x = 1-\cos^2 x\]
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

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

- anonymous

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

- 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

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

- anonymous

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

- zepdrix

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

- anonymous

okay sorry I thought you were speaking to me

- 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

Understand what we did for the first step kenney? :O it's a little tricky, we're taking advantage of a trig identity.

- zepdrix

kenney too quiet -_- hehe

- anonymous

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

- anonymous

but yes i understand

- anonymous

if we go too fast for you tell us okay

- 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

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

- anonymous

how will locating on the unit circle help?

- anonymous

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

- 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

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

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

- 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

what godo? :D

- anonymous

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

- 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

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

- anonymous

ok

- 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

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

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

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

- anonymous

7pi/4?

- zepdrix

yay \:D/

- anonymous

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

- anonymous

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

- 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

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

- anonymous

no worries..all help was appreciated!

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