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anonymous
 one year ago
Find the xcoordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π].
so far I found f'(x)=2cos(2x)+2
cos(2x)=1
anonymous
 one year ago
Find the xcoordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]. so far I found f'(x)=2cos(2x)+2 cos(2x)=1

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myininaya
 one year ago
Best ResponseYou've already chosen the best response.1so far good do you know how to solve: \[\cos(\theta)=1 \text{ for } \theta \]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1have you ever seen the unit circle before?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1Do x coordinates of the pairs represents cos these are the numbers we want to look at can you find when the xcoordinates on the unit circle will be 1?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1The x coordinates * (not do)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1pi is going to be one solution there is another solution we were solving cos(2x)=1 on [0,2pi] but I replaced 2x with theta so we had 0<=x<=2pi and x is theta/2 so 0<=theta/2<=2pi multiply 2 on both sides we have 0<=theta<=4pi so we actually want to solve cos(theta)=1 on [0,4pi] but that isn't too terrible we know pi is one solution in that interval but pi+2pi is another we wanted to solve for x not theta but we know the relationship between x and theta is given by 2x=theta so we have \[2x=\pi \text{ or also } 2x=\pi+2\pi\] simplify and solve for x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that would be 2x=3pi so x=pi/2 and x=3pi/2

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1sounds great to me :)
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