How to find all solutions for x in the interval [0, 2pi) for 2cos(3x) - 1 = 0?

[lovemake1]

Homework Statement

Find all solution for x in the interval [0, 2pi)
2cos(3x) - 1 = 0

Homework Equations

The Attempt at a Solution

2cos3x = 1
cos3x = 1/2

since x = [0,2pi)
therefore, 3x is [0,6pi)

so.. here is my problem i know that cosx =1/2 is 1/3pi, and how would i go from here?
will i be adding 1/3pi + 2n pi or just 1/3pi + n pi, and how do we know which one to use?
please help,

i will be refreshing the page frequently to respond to your next suggestion or errors i might have.

please help, thanks

 

[rock.freak667]

Homework Statement

Find all solution for x in the interval [0, 2pi)
2cos(3x) - 1 = 0

Homework Equations

The Attempt at a Solution

2cos3x = 1
cos3x = 1/2

since x = [0,2pi)
therefore, 3x is [0,6pi)

so.. here is my problem i know that cosx =1/2 is 1/3pi, and how would i go from here?
will i be adding 1/3pi + 2n pi or just 1/3pi + n pi, and how do we know which one to use?
please help,

First you need to find all the solutions from 0 to 2π. Cosine is also positive in quadrant 4, meaning that another solution is 2π-π/3.

So you solutions will be x = π/3,2π-π,3

now when you go around the circle again, you will need to add 2π to all your solutions. Just go around another time and you should get all of your solutions for the problem.

 

[lovemake1]

so our goal is to find every positive solution ? meaning for cos = 1/2, quadrant I and IV?
please correct me on this one,
and if it was cos = -1/2 we would find answers from quadrant II and III
as n - 1/2 and n + 1/2, please let me know if my understanding is correct.

Thanks

 

[rock.freak667]

so our goal is to find every positive solution ? meaning for cos = 1/2, quadrant I and IV?
please correct me on this one,
and if it was cos = -1/2 we would find answers from quadrant II and III
as n - 1/2 and n + 1/2, please let me know if my understanding is correct.

Thanks

I am not sure what you mean by 'n+1/2', but the part about cosine being negative is correct.

 

[lovemake1]

i've found all the solutions for cos = 1/2

pi/3, 2pi/3, 7pi/3, 8pi/3, 13pi/3, 14pi/3, 19pi/3 ( this is greater than 6pi, so excluded)

but since this is 3x not x, which extends the interval from [0,2pi) to [0,6pi]
would we just divide all our answers by 3?

 

[eumyang]

i've found all the solutions for cos = 1/2

pi/3, 2pi/3, 7pi/3, 8pi/3, 13pi/3, 14pi/3, 19pi/3 ( this is greater than 6pi, so excluded)

but since this is 3x not x, which extends the interval from [0,2pi) to [0,6pi]
would we just divide all our answers by 3?

The ones I bolded above are wrong. Those angles are in Quadrant II, not IV. Try again.

Once you fix the above, then yes, you would then have to divide all of the answers by 3.

 

[lovemake1]

right, 2pi/3 shud be 5pi/3
and 8pi/3 shud be 11pi/3
14pi/3 shud be 17pi/3

 

[joshiemen]

Homework Statement

Find all solution for x in the interval [0, 2pi)
2cos(3x) - 1 = 0

Homework Equations

The Attempt at a Solution

2cos3x = 1
cos3x = 1/2

since x = [0,2pi)
therefore, 3x is [0,6pi)

so.. here is my problem i know that cosx =1/2 is 1/3pi, and how would i go from here?
will i be adding 1/3pi + 2n pi or just 1/3pi + n pi, and how do we know which one to use?
please help,

i will be refreshing the page frequently to respond to your next suggestion or errors i might have.

please help, thanks

Shud not the highlighted be [0,2pi/3)?

 

[rock.freak667]

Shud not the highlighted be [0,2pi/3)?

No, that would mean you are solving cos(x/3)=1/2

If 0≤x<2π, then 0≤3x<6π

 

[joshiemen]

No, that would mean you are solving cos(x/3)=1/2

If 0≤x<2π, then 0≤3x<6π

Dear Rock.freak667,

Does not period of a cosine function

y = a cos kx (k>0)

complete one period as kx varies from 0 to 2pi, that's 0<=kx<=2pi or for 0<=x<=2pi/k? For example period of cos 3x is "2pi/3"...

May be I am mixing up two separate things, ...pls explain, eager to understand...

Thanks,

Josh.