Question

....

Answer 1

what is our derivative of f wrt.x?

Answer 2

2-2cos(2x)

Answer 3

+ not -

factor out the 2 and ... what do you see that we can do then?

Answer 4

or set it equal to 0, subtract off the 2, divide by -2 etc ...

Answer 5

You get 2(1-cosx)

Answer 6

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

Answer 7

when does 1+cos(2x) = 0?

Answer 8

At 0?

Answer 9

no cos(0) = 1, not -1

Answer 10

cos(2x) = -1 is what we want for 1 + cos(2x) to be zero

Answer 11

cos(pi) = -1, so when does 2x = pi?

Answer 12

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

Answer 13

Thats the answer?

Answer 14

those are approaches that I would take to find the answer ...

Answer 15

Which do you think is the best approach

Answer 16

ive only given 1 approach

Answer 17

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.

Answer 18

2x= pi, and 2x = 3pi

Answer 19

** our interval is x=0 to 2pi ...

Answer 20

So x=pi/2 and x=3/2pi

Answer 21

correct

Answer 22

Thank you!!!