Question

Please Help. Will give medal.

Answer 1

Find all solutions to the equation in the interval [0, 2π)
cos 2x - cos x = 0

Answer 2

if that's cos(2x), then use the identity

cos(2x) = cos^2(x) - sin^2(x)

Answer 3

So cos^2(x)-sin^2(x)-cos(x)=0?

Answer 4

Then you can use sin^2(x) = 1 - cos^2(x)

Answer 5

Yeah, just though of that...lol

Answer 6

So what do we do next?

Answer 7

cos^2(x)-sin^2(x)-cos(x)=0

cos^2(x)-(1-cos^2(x))-cos(x)=0

cos^2(x)-1+cos^2(x)-cos(x)=0

2cos^2(x)-cos(x)-1=0

Answer 8

let z = cos(x), so z^2 = cos^2(x) giving you

2z^2 - z - 1 = 0

Answer 9

solve for z, then use that to find x

Answer 10

z=1 and z=-1/2?

Answer 11

use that to find x

Answer 12

so because z = cos(x), and z = 1, we know

z = 1

cos(x) = 1

x = ??

Answer 13

x is 1? since x is cos?

Answer 14

Or am I mixing things up?

Answer 15

x isn't 1

Answer 16

you might be

Answer 17

1(2) = 1^2 - sin^2(x)

Answer 18

2=1-sin^2(x)

Answer 19

use the unit circle to solve cos(x) = 1

Answer 20

I mean 0

Answer 21

or 2pi?

Answer 22

they both work, but only x = 0 is in the interval [0, 2π)

Answer 23

now solve cos(x) = -1/2

Answer 24

x=2/3pi pi (n)ez?

Answer 25

ops

Answer 26

It should be half way between 180 and 90

Answer 27

degrees

Answer 28

since cosine is negative, you should be in quadrants II and III

Answer 29

270? and..

Answer 30

90?

Answer 31

pi/2 and 3pi/2?

Answer 32

cos(x) = -1/2, not cos(x) = 0

Answer 33

lol, whats wrong with me, let me look at this again

Answer 34

120 degrees!

Answer 35

so.. its...

Answer 36

what else

Answer 37

umm 300 degrees?

Answer 38

Oh thats wrong

Answer 39

if cos(theta) < 0, then theta is in quadrants II and III

Answer 40

I know this

Answer 41

Its 24o degrees

Answer 42

*240

Answer 43

good

Answer 44

^_^ yess!

Answer 45

so x = 0, 120, 240

Answer 46

0, 2pi/3 and 4pi/3

Answer 47

?

Answer 48

Yeah, that should be it. Thankyou! *Gives virtual hug*