Question

solve equation on the interval [0,2pi)

cos^2(x)=cosx

dont know what todo, need help where to start please.

Answer 1

\[\cos^2(x)-\cos(x)=\cos(x)(\cos(x)-1)=0 => \cos(x)=0 or \cos(x)-1=0\]

Answer 2

hello myininaya and imran

Answer 3

hey

Answer 4

i'm so busy
i'm not use to this

Answer 5

site is freezing up

Answer 6

you can treat this as a regular quadratic equation

Answer 7

well the answerkey given is in radians,

Answer 8

but im lost atm,

Answer 9

yup that right, what you would do is simply: cos^2x-cosx=0
Then factor out a cos and get: cox(x)(cos(x)-1)=0 now set each term to be zero

Answer 10

cos(x)=0 when x=pi/2 or 3pi/2

cos(x)-1=0 when cos(x)=1
cos(x)=1 when x=0

Answer 11

so cosx=0 and cosx=1???

Answer 12

yup that right

Answer 13

answer is {pi/4, 3pi/4, 5pi/4, & 7pi/4} i think i use the unit circle right?

Answer 14

since its restriction is (o,2pi)

Answer 15

myininaya has the correct answer

Answer 16

how did you get that solution set?

Answer 17

yes, so within that interval you determine when cos x is 0 and 1

Answer 18

within the interval [0,pi], myin has it right

Answer 19

within the interval [0,2pi) you mean

Answer 20

oh ok thank you, im starting to see it

Answer 21

yes, sorry i meant [0,pi). Thank you for the correction

Answer 22

lol 2pi*

Answer 23

lol

Answer 24

its okay lagrange we know what you mean

Answer 25

can you tell i am having a horrible day, lol

Answer 26

OOPPS wrong asnwer key....lol

its (0,pi/2, and 3pi/2) sorry now i get it haha