Find the values of x on the curve below at which the tangent is horizontal. Use n as an arbitrary integer. (Enter your answers as a comma-separated list.)

y= (sin(x))/ (2+cos(x))

When the tangent is horizontal, dy/dx = 0

Have you taken the derivative?
Set it equal to zero and solve for x

I have, but I keep getting an incorrect final answer.

It would be cosx/-sinx, right?

Oh no!

You can't just differentiate top and bottom of a fraction, you must use the quotient rule
y = (sin(x))/ (2+cos(x))
dy/dx = ( (2+cosx)(cosx) - sinx(-sinx)/(2+cosx)^2
= (2cosx + cos^2 x + sin^2x)/(2+cosx)^2
= (2cosx + 1)/(2 + cosx)^2
= 0
then 2cosx + 1 = 0
cosx = -1/2
x = 2π/3 or x = 4π/3

To find the values of x on the curve at which the tangent is horizontal, we need to determine the points where the derivative of the curve with respect to x is equal to zero.

First, let us find the derivative of the curve y = (sin(x))/(2 + cos(x)). We can use the quotient rule to find the derivative of the function.

Step 1: Find the derivative of the numerator
dy/dx (sin(x)) = cos(x)

Step 2: Apply the chain rule to the denominator
dy/dx (2 + cos(x)) = -sin(x)

Now, let's find the derivative of the function using the quotient rule:

dy/dx = [(cos(x) * (2 + cos(x))) - ((sin(x)) * (-sin(x))) / (2 + cos(x))^2

Simplifying further:

dy/dx = (2cos(x) + cos^2(x) + sin^2(x)) / (2 + cos(x))^2
dy/dx = (2cos(x) + 1) / (2 + cos(x))^2

To find where the tangent is horizontal, we need to set the derivative equal to zero:

0 = (2cos(x) + 1) / (2 + cos(x))^2

To solve this equation, we can start by cross-multiplying:

0 * (2 + cos(x))^2 = 2cos(x) + 1

Expanding the square:

0 = 4 + 4cos(x) + cos^2(x) - 2cos(x) - 1

0 = cos^2(x) + 2cos(x) + 3

Rearranging the equation:

cos^2(x) + 2cos(x) + 3 = 0

Now, you can solve this quadratic equation for cos(x) by factoring, using the quadratic formula, or completing the square. However, it is important to note that the range of the cosine function is -1 to 1, so there are no real solutions for cos(x) in this equation.

Therefore, there are no values of x on the curve where the tangent is horizontal.