Question

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

Answer 1

you there

Answer 2

yes

Answer 3

okay tell me what you know first?

Answer 4

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

Answer 5

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

Answer 6

yes!

Answer 7

recall the untit circle and where sin is and so forth

Answer 8

its the y coordinate

Answer 9

yes

Answer 10

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?

Answer 11

im not sure...

Answer 12

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

Answer 13

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

Answer 14

can you simplify it?

Answer 15

yes

Answer 16

to what? sorry

Answer 17

one sec

Answer 18

@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 :)

Answer 19

yeah I could use some help here I am tying to think how to do this in a simple way can you retrice me?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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 :)

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

okay sorry I thought you were speaking to me

Answer 28

Using this identity:\[\large \sin^2 x = (1-\cos^2 x)\]
Our equation becomes:\[\large 4\sin^2 x +4\sqrt2 \cos x - 6 = 0\]\[\large 4(1-\cos^2 x) +4\sqrt2 \cos x - 6 = 0\]

Answer 29

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

Answer 30

kenney too quiet -_- hehe

Answer 31

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

Answer 32

but yes i understand

Answer 33

if we go too fast for you tell us okay

Answer 34

answer choices are:
a. 3π/4, 7π/4
b. π/4, 7π/4
c. π/4, 5π/4
d. 3π/4, 5π/4

Answer 35

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

Answer 36

how will locating on the unit circle help?

Answer 37

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

Answer 38

\[\large 4(1-\cos^2 x) +4\sqrt2 \cos x - 6 = 0\]
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.
\[\large 4u^2-4\sqrt2u+2=0\]
\[\large u=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 39

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

Answer 40

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

Answer 41

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.

Answer 42

what godo? :D

Answer 43

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

Answer 44

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.

Answer 45

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

Answer 46

ok

Answer 47

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

Answer 48

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.

Answer 49

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??

Answer 50

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

Answer 51

7pi/4?

Answer 52

yay \:D/

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

no worries..all help was appreciated!