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)

B.-2,3 because it is touch the -3 i think that the answer