Find the x-coordinates where f ' (x) = 0 for f (x) = 2x - sin (2x) in the interval [0, 2π].

The x-coordinates where f ' (x) = 0 for f (x) = 2x - sin (2x) in the interval [0, 2π] are: x=0, π, and 2π.

Step-by-step explanation:

f (x) = 2x-sin (2x)

f' (x) = [f (x) ]'

f' (x) = [2x-sin (2x) ]'

f' (x) = (2x) '-[sin (2x) ]'

f' (x) = 2-cos (2x) (2x) '

f' (x) = 2-cos (2x) (2)

f' (x) = 2-2 cos (2x)

f' (x) = 0, then:

2-2 cos (2x) = 0

Solving for x. First solving for cos (2x) : Subtracting 2 from both sides of the equation:

2-2 cos (2x) - 2=0-2

-2 cos (2x) = -2

Dividing both sides of the equation by - 2:

-2 cos (2x) / (-2) = -2 / (-2)

cos (2x) = 1

If cos (2x) = 1, then:

2x=2nπ

Dividing both sides of the equation by 2:

2x/2=2nπ/2

x=nπ

If n=-1→x=-1π→x=-π, and - π is not in the interval [0,2π], then x=-π is not in the solution.

If n=0→x=0π→x=0, and 0 is in the interval [0,2π], then x=0 is in the solution.

If n=1→x=1π→x=π, and π is in the interval [0,2π], then x=π is in the solution.

If n=2→x=2π, and 2π is in the interval [0,2π], then x=2π is in the solution.

If n=3→x=3π, and 3π is not in the interval [0,2π], then x=2π is not in the solution.

Solution: x=0, π, and 2π.