anonymous
  • anonymous
2 cos^2 x - 3 cos x + 1 = 0 for 0 is less than or equal to x is less than 2pi
Mathematics
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anonymous
  • anonymous
2 cos^2 x - 3 cos x + 1 = 0 for 0 is less than or equal to x is less than 2pi
Mathematics
katieb
  • katieb
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anonymous
  • 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
  • anonymous
yea the first one
anonymous
  • 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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anonymous
  • anonymous
yea i can
anonymous
  • anonymous
what is/are the solutions ? y = ???
anonymous
  • anonymous
cos2x=2cos^2(x)-1
anonymous
  • anonymous
2(2cos^2(x)-1)-3cosx+1=0 -> 4cos^2(x)-3cosx-1=0 -> (4cosx+1)(cosx-1)=0
anonymous
  • anonymous
x1 = -0.5 x2 = -1
anonymous
  • anonymous
so set 4cosx+1=0 or cos-1 =0
anonymous
  • anonymous
hmmm.. i got the same but POSITIVE values... i'm double checking...
anonymous
  • 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
  • anonymous
ok so far?
anonymous
  • anonymous
yea
anonymous
  • anonymous
you forgot to x2
anonymous
  • anonymous
it is 2cos2x
anonymous
  • anonymous
it is 2 cos^2 x - 3
anonymous
  • anonymous
oh i thought it is cos2x
anonymous
  • anonymous
since y=cosx, we have these two equations: \( \large cosx=\frac{1}{2}\) and \(\large cosx=-1 \) can you solve these?
anonymous
  • anonymous
oops... that second one should be POSITIVE one...
anonymous
  • anonymous
do you use the unit circle?
anonymous
  • anonymous
yes i do
anonymous
  • 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
  • anonymous
@joshi its 2 cos^2 x -3 i made a mistake in typing it
anonymous
  • 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
  • anonymous
ohhh ok that makes more sense haha
anonymous
  • anonymous
yeah... i got a clarification from the asker with my first question... :)
anonymous
  • anonymous
60 degrees?
anonymous
  • anonymous
|dw:1350069794735:dw| yep... 60 degrees is one of 'em....
anonymous
  • anonymous
or pi/3
anonymous
  • anonymous
you got the second angle?
anonymous
  • anonymous
and 45 degrees?
anonymous
  • 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
  • anonymous
|dw:1350070145499:dw|
anonymous
  • anonymous
or take a look at page 3: http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf
anonymous
  • anonymous
240 degrees?
anonymous
  • anonymous
240 degrees is in the third quadrant... this is the fourth quadrant: |dw:1350070347242:dw|
anonymous
  • anonymous
ohh ok 300 degrees..
anonymous
  • anonymous
yes... :) cos(60 degrees) = 1/2 and cos(300 degrees) = 1/2 now solve the other equation: cos x = 1
anonymous
  • anonymous
and remember, \(\large 0\color {red}{\le} x<2\pi \)
anonymous
  • anonymous
there should be only one solution for this one....
anonymous
  • anonymous
using the unit circle, what angle gives you an x-coordinate of 1?
anonymous
  • anonymous
360 degrees?
anonymous
  • 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
  • anonymous
HINT: \(\huge \le \)
anonymous
  • anonymous
ohhh yea it will 180 degrees
anonymous
  • anonymous
no.. cos(180 degrees) = -1 try again....
anonymous
  • anonymous
cos x = 1 where \( \large 0\color {red}{\le} x<2\pi\)
anonymous
  • anonymous
3pi/2?
anonymous
  • anonymous
what about the left side of \(\large 0\color {red}{\le} x<2\pi \) ???? x = 0 ? cos 0 = ???
anonymous
  • anonymous
ok... let's put it this way... what's an angle that is coterminal with 360 degrees?
anonymous
  • anonymous
positive 360 and negative -360
anonymous
  • 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
  • anonymous
so it will be x=1.823 radian

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