Question

cos^2x+2cosx+1=0

Answer 1

\[\cos^2(x) +2\cos(x) +1=0\]\[\text{let a}= \cos(x)\]\[a^2 +2a+1=0\]What two numbers multiply to 1 and add to give 2?

Answer 2

1 and 1

Answer 3

Awesome, so that would become : \((a+1)(a+1)=0\) or otherwise written as \((a+1)^2 = 0\)

Answer 4

so (cosx+1)^2=0?

Answer 5

Yes :)

Answer 6

then how do I factor that?

Answer 7

Take the square root of both sides,

Answer 8

so it becomes cosx+1=0?

Answer 9

Mmhmm.

Answer 10

Now we subtract by -1 on both sides of the equation.

Answer 11

cosx=-1

Answer 12

Alright, and now we isolate x by taking the inverse on both sides of the equation.

Answer 13

so x=cos^-1(-1)?

Answer 14

mmhmm. so we find where \(\cos(\theta) = -1\) the first time around the unit circle, where will that be?

Answer 15

180 degrees

Answer 16

or \(\pi\).

Answer 17

then we know cosine has a period of \(2\pi\), so every time we go around our unit circle, we will add \(\pi\) to our solution, therefore \[\boxed{x=\pi +2k\pi}\]

Answer 18

Does that make sense? :)

Answer 19

yeah, is that all there is?

Answer 20

Yep :D

Answer 21

I don't have to plug any numbers in? it says "solve using interval notation and list all angles that fit in the interval (0,2pi)?

Answer 22

Ahh.. ok.

Answer 23

Would only plugging in 0 and -1 work to fit the interval?

Answer 24

to get x=pi and x=-pi?

Answer 25

One sec, writing this up :P

Answer 26

well, let's go back to \[\cos(x) = -1\]That means x = \(-\pi\) since that is the only place where \(\cos(x) = -1\) in the interval \(0 < x < 2\pi\)

Answer 27

okay I got it. thank you!

Answer 28

Haha, no problem ! Sorry, these are not my forte :\