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looks good to me
i never remember the formula, so i set \(2x-\frac{\pi}{3}=0\) and solve for \(x\)

|dw:1349892894075:dw|

i meant set \[2x-\frac{\pi}{3}=0\] not
\[2x-\frac{x}{3}=0\]

haha okay, i thought = to zero

whew

haha sorry about that

okay so to figure the points, i start at pi/6 and move right by pi/6?

lets go slow because i do not want to confuse you

you know \(\cos(0)=1\) right?

OH! i forgot i have to move it up 1

okay so then i have it like this:
|dw:1349893756732:dw|

then i move up 1

|dw:1349893880189:dw|

|dw:1349894136637:dw|

how did you do that?

you know the period is pi, so every time you increase x by pi you will get the same value as x

sorry, im new to precalculus. all this is super new to me

the hight point i got is (pi/6,2) but how do i get (7pi/6, 2)?
pi/6 + pi = pi/6 + 6pi/6 = 7pi/6

how did you do that?
are you asking how to get the lowest point at x= 2/3 pi ?

|dw:1349897328597:dw|

this is the graph i got, but i got pi/2 as my lowest.
did i do something wrong?

type
y = cos (2x - pi/3) +1
in the google search window, it should graph it

okay im lookin at the graph

so how do i figure the lowest with that graph?

cos(2x-pi/3) = -1 okay i get this like u did

i end up getting 2pi/3.
and that graph on google is confusing me with different numbers

*The graph shows the curve with x in radians.