onegirl
2 cos^2 x - 3 cos x + 1 = 0 for 0 is less than or equal to x is less than 2pi
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dpaInc
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the first term with the cosine... is it:
\(\large 2cos^2x \) or \(\large 2cos(2x) \)
i'm thinking it's the first one?
onegirl
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yea the first one
dpaInc
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think of it this way..
let y=cosx
so your equation becomes: \(\large 2y^2-3y+1=0 \)
can you solve this quadratic?
onegirl
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yea i can
dpaInc
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what is/are the solutions ?
y = ???
z3529080
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cos2x=2cos^2(x)-1
z3529080
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2(2cos^2(x)-1)-3cosx+1=0 -> 4cos^2(x)-3cosx-1=0 -> (4cosx+1)(cosx-1)=0
onegirl
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x1 = -0.5 x2 = -1
z3529080
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so set 4cosx+1=0 or cos-1 =0
dpaInc
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hmmm.. i got the same but POSITIVE values... i'm double checking...
dpaInc
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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
dpaInc
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ok so far?
onegirl
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yea
z3529080
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you forgot to x2
z3529080
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it is 2cos2x
onegirl
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it is 2 cos^2 x - 3
z3529080
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oh i thought it is cos2x
dpaInc
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since y=cosx, we have these two equations:
\( \large cosx=\frac{1}{2}\) and \(\large cosx=-1 \)
can you solve these?
dpaInc
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oops... that second one should be POSITIVE one...
dpaInc
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do you use the unit circle?
onegirl
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yes i do
joshi
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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?
onegirl
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@joshi its 2 cos^2 x -3 i made a mistake in typing it
dpaInc
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ok... so look at what angle gives you a cosine of 1/2....
HINT: there are two of 'em from 0 to 2pi
joshi
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ohhh ok that makes more sense haha
dpaInc
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yeah... i got a clarification from the asker with my first question... :)
onegirl
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60 degrees?
dpaInc
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|dw:1350069794735:dw|
yep... 60 degrees is one of 'em....
onegirl
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or pi/3
dpaInc
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you got the second angle?
onegirl
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and 45 degrees?
dpaInc
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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...
dpaInc
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|dw:1350070145499:dw|
onegirl
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240 degrees?
dpaInc
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240 degrees is in the third quadrant... this is the fourth quadrant:
|dw:1350070347242:dw|
onegirl
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ohh ok 300 degrees..
dpaInc
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yes... :)
cos(60 degrees) = 1/2 and cos(300 degrees) = 1/2
now solve the other equation: cos x = 1
dpaInc
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and remember, \(\large 0\color {red}{\le} x<2\pi \)
dpaInc
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there should be only one solution for this one....
dpaInc
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using the unit circle, what angle gives you an x-coordinate of 1?
onegirl
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360 degrees?
dpaInc
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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?
dpaInc
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HINT: \(\huge \le \)
onegirl
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ohhh yea it will 180 degrees
dpaInc
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no..
cos(180 degrees) = -1
try again....
dpaInc
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cos x = 1
where \( \large 0\color {red}{\le} x<2\pi\)
onegirl
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3pi/2?
dpaInc
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what about the left side of \(\large 0\color {red}{\le} x<2\pi \) ????
x = 0 ?
cos 0 = ???
dpaInc
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ok... let's put it this way...
what's an angle that is coterminal with 360 degrees?
onegirl
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positive 360 and negative -360
dpaInc
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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.
onegirl
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so it will be x=1.823 radian