Calculus of a Single Variable, Chapter 2, 2.2, Section 2.2, Problem 62

Take the derivative of the function.
y' = sqrt3 -2sin(x)
Set the derivative function equal to zero, since we want to know where x is when the slope is zero.
0 = sqrt3 -2sin(x)
2sin(x)=sqrt3
sin(x)=sqrt(3) /2
We need to find all the angles of x where the y-coordinate of the unit circle equals to \sqrt3 /2 on the domain [0,2pi).
The y-value sqrt3 / 2 will exist in the first and second quadrants of the unit circle.
The angles are:
x=pi/3,(2pi)/3
These two values will give us zero when we substitute them back into the derivative function.
To get the points, plug these two values back to the original function.

y=sqrt3 (pi/3) + 2cos(pi/3)= sqrt3 (pi/3) +1
The first point is: (pi/3, (sqrt3 pi)/3 +1)
y=sqrt3 ((2pi)/3) + 2(cos((2pi)/3))= (2sqrt3 pi) /3 + 2(-1/2)= (2sqrt3 pi) /3-1
The second point is: ((2pi)/3,(2sqrt3 pi) /3-1)