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what is our derivative of f wrt.x?
2-2cos(2x)
+ not - factor out the 2 and ... what do you see that we can do then?

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or set it equal to 0, subtract off the 2, divide by -2 etc ...
You get 2(1-cosx)
2(1 + cos(2x)) = 0 since 2 is never 0, that only leaves 1+cos(2x) to be zero for this to have any chance of working
when does 1+cos(2x) = 0?
At 0?
no cos(0) = 1, not -1
cos(2x) = -1 is what we want for 1 + cos(2x) to be zero
cos(pi) = -1, so when does 2x = pi?
cos(3pi) = -1 as well, any odd multiple of pi makes cos go to -1 2x = 3pi should fit in the interval if i see it correctly
Thats the answer?
those are approaches that I would take to find the answer ...
Which do you think is the best approach
ive only given 1 approach
given that our interval is from x=0 to pi 2x gives us a range from 0 to 4pi to play with cos(pi) and cos(3pi) are both equal to -1 so we will have 2 solutions.
2x= pi, and 2x = 3pi
** our interval is x=0 to 2pi ...
So x=pi/2 and x=3/2pi
correct
Thank you!!!

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