anonymous
  • anonymous
which value of theta satisfies the equation 2cos^2theta - cos theta=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[cos^2(\theta)-cos(\theta)=0\] \[cos(\theta)(cos(\theta)-1)=0\] \[cos(\theta)=0\] or \[cos(\theta)=1\]
anonymous
  • anonymous
oooooooops forgot the two!
anonymous
  • anonymous
ignore my post, or just modify with the two.

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More answers

anonymous
  • anonymous
is there a formula to memorize forthis?
anonymous
  • anonymous
\[cos(\theta)(2cos(\theta)-1)=0\] \[cos(\theta)=0\] \[cos(\theta)=\frac{1}{2}\]
anonymous
  • anonymous
oh no the first part has nothing at all to do with trig
myininaya
  • myininaya
costheta(2costheta-1)=0 cost=0 => t=pi/2,3pi/2,... 2npi+pi/2 and 2npi+3p/2 for n=0,1,2,... 2cost-1=0 => cost=1/2 => t=pi/3,11pi/6,.... 2npi+pi/3 and 2npi+11pi/6 for n=0,1,2..
anonymous
  • anonymous
like solving \[2x^2-x=0\] \[x(2x-1)=0\] \[x=0\] \[x=\frac{1}{2}\]
anonymous
  • anonymous
oh no im confused
anonymous
  • anonymous
the trig part comes now.
myininaya
  • myininaya
oops that 11pi/6 is wrong its 5pi/3
anonymous
  • anonymous
the first job is to solve for \[cos(\theta)\] the next step is to solve for \[\theta\]
anonymous
  • anonymous
the first steps have nothing at all to do with trig. no formulas, no nothing.
anonymous
  • anonymous
ok
anonymous
  • anonymous
think of \[cos(\theta)\] as x and solve for x. you have \[2x^2-x=0\] and this is easy to solve
anonymous
  • anonymous
we get \[x=0\] or \[x=\frac{1}{2}\] so now \[cos(\theta)=0\] or \[cos(\theta)=\frac{1}{2}\]
anonymous
  • anonymous
the trig part comes now, seeing if we can find numbers for theta that will make this work.
anonymous
  • anonymous
can you do the next step, that is, can you find theta?
anonymous
  • anonymous
im not sure/
anonymous
  • anonymous
isnt it 0?
anonymous
  • anonymous
ok so first off we want to find where \[cos(\theta)=0\]
anonymous
  • anonymous
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf
myininaya
  • myininaya
just look between 0 and 2pi first to see where cosine is 0
anonymous
  • anonymous
best cheat sheet. look on it to see where cosine is 0. it is not at 0, because \[cos(0)=1\]
myininaya
  • myininaya
then after just add 2npi for n=0,1,2,3,... blah blah
anonymous
  • anonymous
cos is the x correct?
anonymous
  • anonymous
yes we already know x=0, so therefore \[cos(\theta)=0\] now we are looking for \[\theta\] that is we are looking for an angle (number) whose cosine is 0
myininaya
  • myininaya
what happens at pi/2?
anonymous
  • anonymous
60 defgrees
anonymous
  • anonymous
the very bottom of the cheat sheet i sent has all the angles associated with the unit circle, and all the coordinates. look on the unit circle to see where the first coordinate is 0
anonymous
  • anonymous
0,1
anonymous
  • anonymous
90
anonymous
  • anonymous
yay
anonymous
  • anonymous
if you are working in degrees it is 90
anonymous
  • anonymous
radians \[\frac{\pi}{2}\]
myininaya
  • myininaya
is there another between 0 and 2pi, that cosine will be zero?
anonymous
  • anonymous
but on my wksht theres noone of those answers.
anonymous
  • anonymous
yes at 270
anonymous
  • anonymous
whatever you prefer. so that takes care of \[cos(\theta)=0\] for the moment. now we need \[cos(\theta)=\frac{1}{2}\]
anonymous
  • anonymous
now that one is 60
anonymous
  • anonymous
yes 90 works, and 270 also works
anonymous
  • anonymous
or 300 degrees
anonymous
  • anonymous
exactly
myininaya
  • myininaya
or 3pi/2 now we can get cosine is 0 when we start at these angles and go around the circle again (2pi) so the answer for cos(t)=0 is t=2npi+pi/2 and t=2npi+3pi/2
anonymous
  • anonymous
you are working in degrees so you have the answers.
myininaya
  • myininaya
that whole paragraph is still talking about the first one k?
anonymous
  • anonymous
theo nly answers available are pi over 6 0 pi/4 or pi/m3
anonymous
  • anonymous
then they are missing \[\frac{\pi}{2}\]
anonymous
  • anonymous
\[\frac{\pi}{3}\] works, from the cheat sheet
anonymous
  • anonymous
if those are your possible answers you are working in radians not degrees.
anonymous
  • anonymous
\[2 \cos^2 \theta- \cos \theta = 0\]
anonymous
  • anonymous
i know but i like using degrees better.
anonymous
  • anonymous
so was pi/3 the answer? vecause cos was 1/2?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i semi get it,thank you
anonymous
  • anonymous
i guess it just asked 'which one of these is a solution"
anonymous
  • anonymous
did you look at the cheat sheet? because it is a good one
anonymous
  • anonymous
yeah which value satisfies the the equation
anonymous
  • anonymous
yes, it had a lot of good material thank you
anonymous
  • anonymous
could you helo me with my other math questions?
l
  • l
2 cos_th^2 - cos_th = 0 cos_th (2 cos_th - 1) = 0 -> cos_th =0 or 1/2 -> th = theta = +pi/2 or - pi/2 or -pi/3 or +pi/3 *
anonymous
  • anonymous
@LynFran

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