# `y = x + sin(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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We will need to take the derivative of the function and set the derivative equal to zero.

`y' = 1+cos(x)`

`0 = 1+cos(x)`

`cos(x)= -1`

The value of `x` is the angle where we have a point on the unit circle with an x coordinate value of -1.

The domain given exists from `[0,2 pi)` .

The only value that the unit circle will have a `(-1,0)` point is when the angle is `pi` radians, or 180 degrees.

We can work our way backward and find that in both radians and degrees ,respectively:

`cos(pi)=cos(180) = -1`

Therefore, the value of `x` in radians where we have a slope of zero is:

`x=pi`

Substitute this angle back to the original function to find the point.

`y=x+sin(x)`

`y=pi+sin(pi)`

The value of sine at `pi` radians or 180 degrees is 0.

`y=pi+0= pi`

The exact point existent on the given domain would be:

`(pi,pi)`