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