Find the points 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

did you mean

y = sinx/(2+cosx) ??

I will assume you did.

dy/dx = ( (2+cosx)(cos) - sinx(-sinx) )/(2+cosx)^2
= (2cosx + cos^2 x + sin^2 x)/(2+cosx)^2
= (1 + 2cosx)/(2+cosx)^2
for a horizontal tangent, dy/dx = 0
or
1 + 2cosx = 0
cosx = -1/2
x = 2π/3 or x = 4π/3 , (120° or 240° )
the period of cosx = 2π

so general solutions:
x = 2π/3 + n(2π) or
x = 4π/3 + n(2π)

Well, well, well, finding the points where the tangent is horizontal, huh? Sounds like you're looking for some "flat" out answers! Let's dive into it.

To find where the tangent is horizontal on the curve y = (sin x)/2 + cos x, we need to find where the derivative of the curve is equal to zero. So, let's find the derivative:

dy/dx = (1/2) cos x - sin x.

Now, we set the derivative equal to zero:

(1/2) cos x - sin x = 0.

Multiplying everything by 2 to get rid of those pesky denominators, we get:

cos x - 2 sin x = 0.

Now, let's use some trigonometric identities to simplify this equation. We know that cos x = cos^2 x + sin^2 x, so we can rewrite the equation as:

cos x - 2 (cos^2 x + sin^2 x) = 0.

Expanding it out, we get:

cos x - 2 cos^2 x - 2 sin^2 x = 0.

Rearranging terms:

-2 cos^2 x - 2 sin^2 x + cos x = 0.

We can also use the identity sin^2 x = 1 - cos^2 x to further simplify the equation:

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

Simplifying it one more time:

-2 cos^2 x - 2 + 2 cos^2 x + cos x = 0.

Combining like terms:

cos x - 2 = 0.

Now, solving for cos x, we find:

cos x = 2.

Wait a minute! Hold the laughter! There's no solution for cos x = 2, my friend. Cosine takes values between -1 and 1, so there are no real solutions for this equation. That means there are no points on this curve where the tangent is horizontal.

I apologize for the disappointing news, but hey, at least you got a chuckle out of this, right? Keep those questions coming, I'm here to clown around and provide answers.

To find the points on the curve where the tangent is horizontal, we need to find the x-coordinates where the derivative of the function is equal to zero.

Let's start by finding the derivative of y with respect to x:

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

To find the derivative, we can use the chain rule:

dy/dx = (1/2)*cos(x/2) - sin(x)

Now let's set the derivative equal to zero and solve for x:

(1/2)*cos(x/2) - sin(x) = 0

Multiplying the equation by 2 to remove the fraction, we have:

cos(x/2) - 2sin(x) = 0

Now we can solve this equation for x.

First, let's isolate the cosine term:

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

Using the identity sin^2(x) + cos^2(x) = 1, we can replace sin^2(x) with 1 - cos^2(x):

cos(x/2) = 2(1 - cos^2(x))

Expanding and rearranging, we get:

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

Now we can solve this quadratic equation for cos(x):

Let's use a substitution, cos(x) = u:
2u^2 + u - 2 = 0

Factoring the quadratic, we have:
(2u - 1)(u + 2) = 0

Setting each factor equal to zero, we find two possible values for u:

2u - 1 = 0 -> u = 1/2
u + 2 = 0 -> u = -2

Now let's solve for x:

cos(x) = 1/2
Using the inverse cosine, we have:
x = arccos(1/2)

cos(x) = -2
This equation has no real solutions since the cosine function only takes values between -1 and 1.

Now let's find the values of x using the inverse cosine function:

x = arccos(1/2)
The inverse cosine of 1/2 is pi/3 and 5pi/3.

So the values of x where the tangent is horizontal are:
x = pi/3 + 2pi*n
x = 5pi/3 + 2pi*n

Therefore, the points (x, y) on the curve where the tangent is horizontal can be written as a comma-separated list:
(pi/3, y), (5pi/3, y), where y = sin(pi/6) + cos(pi/3) and y = sin(5pi/6) + cos(5pi/3).

To find the points on the curve where the tangent is horizontal, we need to determine where the derivative of the function is equal to zero.

Given the function y = (sin x) / 2 + cos x, we will first find the derivative of this function.

The derivative of y with respect to x can be found using the quotient rule and the chain rule:

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

Now we need to set this derivative equal to zero and solve for x:

(2 * cos x - sin x)/2 = 0

Multiplying both sides of the equation by 2:

2 * cos x - sin x = 0

Next, we combine like terms:

- sin x + 2 * cos x = 0

Now, we can solve this equation to find the values of x where the tangent is horizontal.

To solve for x, we can use the trigonometric identity cos x = sin x to rewrite the equation:

- sin x + 2 * sin x = 0

We can simplify further:

sin x = 0

This equation holds true when x = n * π, where n is an arbitrary integer.

Therefore, the points on the curve where the tangent is horizontal are (n * π, (sin n * π) / 2 + cos n * π) for all integer values of n.