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

- anonymous

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

- jamiebookeater

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- anonymous

the first term with the cosine... is it:
\(\large 2cos^2x \) or \(\large 2cos(2x) \)
i'm thinking it's the first one?

- anonymous

yea the first one

- anonymous

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

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## More answers

- anonymous

yea i can

- anonymous

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

- anonymous

cos2x=2cos^2(x)-1

- anonymous

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

- anonymous

x1 = -0.5 x2 = -1

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok so far?

- anonymous

yea

- anonymous

you forgot to x2

- anonymous

it is 2cos2x

- anonymous

it is 2 cos^2 x - 3

- anonymous

oh i thought it is cos2x

- anonymous

since y=cosx, we have these two equations:
\( \large cosx=\frac{1}{2}\) and \(\large cosx=-1 \)
can you solve these?

- anonymous

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

- anonymous

do you use the unit circle?

- anonymous

yes i do

- anonymous

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?

- anonymous

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

- anonymous

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

- anonymous

ohhh ok that makes more sense haha

- anonymous

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

- anonymous

60 degrees?

- anonymous

|dw:1350069794735:dw|
yep... 60 degrees is one of 'em....

- anonymous

or pi/3

- anonymous

you got the second angle?

- anonymous

and 45 degrees?

- anonymous

nope... we're still solving cosx = 1/2 , right?
x=60 degrees or pi/3 is one...
the second solution is in the fourth quadrant...

- anonymous

|dw:1350070145499:dw|

- anonymous

or take a look at page 3:
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

- anonymous

240 degrees?

- anonymous

240 degrees is in the third quadrant... this is the fourth quadrant:
|dw:1350070347242:dw|

- anonymous

ohh ok 300 degrees..

- anonymous

yes... :)
cos(60 degrees) = 1/2 and cos(300 degrees) = 1/2
now solve the other equation: cos x = 1

- anonymous

and remember, \(\large 0\color {red}{\le} x<2\pi \)

- anonymous

there should be only one solution for this one....

- anonymous

using the unit circle, what angle gives you an x-coordinate of 1?

- anonymous

360 degrees?

- anonymous

yes but 2pi is not included in the interval you're looking at..
360 degrees = 2pi
so what other angle gives you an x-coordinate of 1?

- anonymous

HINT: \(\huge \le \)

- anonymous

ohhh yea it will 180 degrees

- anonymous

no..
cos(180 degrees) = -1
try again....

- anonymous

cos x = 1
where \( \large 0\color {red}{\le} x<2\pi\)

- anonymous

3pi/2?

- anonymous

what about the left side of \(\large 0\color {red}{\le} x<2\pi \) ????
x = 0 ?
cos 0 = ???

- anonymous

ok... let's put it this way...
what's an angle that is coterminal with 360 degrees?

- anonymous

positive 360 and negative -360

- anonymous

no.. the angle you want is 0
because cos(0 degrees) = 1
so you have 3 solutions: x=0, 60, 300 degrees
but since the problem was stated in terms of radians, i'd convert those angles.

- anonymous

so it will be x=1.823 radian

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