# `y = sqrt(3)x + 2cos(x), 0<=x<2pi` Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.

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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)`