Find the x-coordinates 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

From there, you simply need algebra and a calculator that works in radians.

Take the inverse cos of both sides to get 2x = arccos (-1)

Then divide both sides by 2 to get x = arccos (-1) / 2

Put that into a calculator and you get π/2. But because your bounds are 0 to 2π, you have to add π your solution to get the solution on the other side of the unit circle, which would be (3π/2).

Now that you have the x value, put (π/2) and (3π/2) into f (x) to get the y coordinate.

f (π/2) = 2 (π/2) + sin (2 (π/2) = π, which means this solution is just (π/2, π)

f (3π/2 = 2 (3π/2) + sin (2 (3π/2) = 3π, which means this solution is (3π/2, 3π)