Question

2 cos^2 x - 3 cos x + 1 = 0 for 0 is less than or equal to x is less than 2pi

Answer 1

the first term with the cosine... is it:
\(\large 2cos^2x \) or \(\large 2cos(2x) \)

i'm thinking it's the first one?

Answer 2

yea the first one

Answer 3

think of it this way..
let y=cosx
so your equation becomes: \(\large 2y^2-3y+1=0 \)
can you solve this quadratic?

Answer 4

yea i can

Answer 5

what is/are the solutions ?
y = ???

Answer 6

cos2x=2cos^2(x)-1

Answer 7

2(2cos^2(x)-1)-3cosx+1=0 -> 4cos^2(x)-3cosx-1=0 -> (4cosx+1)(cosx-1)=0

Answer 8

x1 = -0.5 x2 = -1

Answer 9

so set 4cosx+1=0 or cos-1 =0

Answer 10

hmmm.. i got the same but POSITIVE values... i'm double checking...

Answer 11

yeah... they should be positive:
\(\large 2y^2-3y+1=0 \)
\(\large (2y-1)(y-1)=0 \)

so
2y - 1= 0 gives y=1/2
y-1 = 0 gives y=1

Answer 12

ok so far?

Answer 13

yea

Answer 14

you forgot to x2

Answer 15

it is 2cos2x

Answer 16

it is 2 cos^2 x - 3

Answer 17

oh i thought it is cos2x

Answer 18

since y=cosx, we have these two equations:

\( \large cosx=\frac{1}{2}\) and \(\large cosx=-1 \)

can you solve these?

Answer 19

oops... that second one should be POSITIVE one...

Answer 20

do you use the unit circle?

Answer 21

yes i do

Answer 22

this is probably a stupid question, but i'm a bit confused. you said let y = cosx so what happens to the first term 2cos2x? like what about the 2 infront of the x?

Answer 23

@joshi its 2 cos^2 x -3 i made a mistake in typing it

Answer 24

ok... so look at what angle gives you a cosine of 1/2....
HINT: there are two of 'em from 0 to 2pi

Answer 25

ohhh ok that makes more sense haha

Answer 26

yeah... i got a clarification from the asker with my first question... :)

Answer 27

60 degrees?

Answer 28

yep... 60 degrees is one of 'em....