which value of theta satisfies the equation 2cos^2theta - cos theta=0

- anonymous

which value of theta satisfies the equation 2cos^2theta - cos theta=0

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- anonymous

\[cos^2(\theta)-cos(\theta)=0\]
\[cos(\theta)(cos(\theta)-1)=0\]
\[cos(\theta)=0\]
or
\[cos(\theta)=1\]

- anonymous

oooooooops
forgot the two!

- anonymous

ignore my post, or just modify with the two.

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## More answers

- anonymous

is there a formula to memorize forthis?

- anonymous

\[cos(\theta)(2cos(\theta)-1)=0\]
\[cos(\theta)=0\]
\[cos(\theta)=\frac{1}{2}\]

- anonymous

oh no the first part has nothing at all to do with trig

- 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

like solving
\[2x^2-x=0\]
\[x(2x-1)=0\]
\[x=0\]
\[x=\frac{1}{2}\]

- anonymous

oh no im confused

- anonymous

the trig part comes now.

- myininaya

oops that 11pi/6 is wrong
its 5pi/3

- anonymous

the first job is to solve for
\[cos(\theta)\]
the next step is to solve for \[\theta\]

- anonymous

the first steps have nothing at all to do with trig. no formulas, no nothing.

- anonymous

ok

- anonymous

think of \[cos(\theta)\] as x
and solve for x.
you have
\[2x^2-x=0\] and this is easy to solve

- anonymous

we get
\[x=0\] or \[x=\frac{1}{2}\]
so now \[cos(\theta)=0\] or \[cos(\theta)=\frac{1}{2}\]

- anonymous

the trig part comes now, seeing if we can find numbers for theta that will make this work.

- anonymous

can you do the next step, that is, can you find theta?

- anonymous

im not sure/

- anonymous

isnt it 0?

- anonymous

ok so first off we want to find where
\[cos(\theta)=0\]

- anonymous

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

- myininaya

just look between 0 and 2pi first to see where cosine is 0

- anonymous

best cheat sheet. look on it to see where cosine is 0. it is not at 0, because \[cos(0)=1\]

- myininaya

then after just add 2npi for n=0,1,2,3,... blah blah

- anonymous

cos is the x correct?

- 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

what happens at pi/2?

- anonymous

60 defgrees

- 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

0,1

- anonymous

90

- anonymous

yay

- anonymous

if you are working in degrees it is 90

- anonymous

radians \[\frac{\pi}{2}\]

- myininaya

is there another between 0 and 2pi, that cosine will be zero?

- anonymous

but on my wksht theres noone of those answers.

- anonymous

yes at 270

- anonymous

whatever you prefer. so that takes care of \[cos(\theta)=0\] for the moment. now we need \[cos(\theta)=\frac{1}{2}\]

- anonymous

now that one is 60

- anonymous

yes 90 works, and 270 also works

- anonymous

or 300 degrees

- anonymous

exactly

- 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

you are working in degrees so you have the answers.

- myininaya

that whole paragraph is still talking about the first one k?

- anonymous

theo nly answers available are pi over 6 0 pi/4 or pi/m3

- anonymous

then they are missing \[\frac{\pi}{2}\]

- anonymous

\[\frac{\pi}{3}\] works, from the cheat sheet

- anonymous

if those are your possible answers you are working in radians not degrees.

- anonymous

\[2 \cos^2 \theta- \cos \theta = 0\]

- anonymous

i know but i like using degrees better.

- anonymous

so was pi/3 the answer? vecause cos was 1/2?

- anonymous

yes

- anonymous

i semi get it,thank you

- anonymous

i guess it just asked 'which one of these is a solution"

- anonymous

did you look at the cheat sheet? because it is a good one

- anonymous

yeah which value satisfies the the equation

- anonymous

yes, it had a lot of good material thank you

- anonymous

could you helo me with my other math questions?

- 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

@LynFran

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