Find the x-coordinates of all points on the curve

f(x) = sin 2x − 2 sin x
at which the tangent line is horizontal. (Enter your answers as a comma-separated list. Use n to represent any integer.)

tangent is horizontal when f'(x) = 0

f'(x) = 2cos2x - 2cosx
f'(x) = 0 when x = 2k*pi or 2k*pi +/- 2pi/3
for all integers k

Well, finding the points where the tangent line is horizontal means finding where the derivative of the function is equal to zero. So, let's take the derivative of f(x) first.

The derivative of f(x) = sin(2x) - 2sin(x) is:

f'(x) = 2cos(2x) - 2cos(x)

Now, we need to find where f'(x) = 0. Let's set it equal to zero and solve for x:

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

When we simplify this equation, we get:

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

Hmm, solving this equation algebraically might not be easy. But don't worry, I'm here to make it fun!

So, let's think outside the box. Since we're dealing with a sine and cosine function, we know that the maximum and minimum values occur at certain angles. For cosine, the maximum value is 1 and the minimum value is -1.

Now, let's think about it. When does cosine become equal to 1 or -1? Well, at x = 0, 2π, 4π, and so on, cosine is equal to 1. And at x = π, 3π, 5π, and so on, cosine is equal to -1.

So, we can say that cos(2x) - cos(x) = 0 when:

2x = 2πn or 2x = π + 2πn

Now, let's solve for x:

When 2x = 2πn:
x = πn

And when 2x = π + 2πn:
x = (π + 2πn) / 2
x = π/2 + πn

So, the x-coordinates where the tangent line is horizontal are πn and π/2 + πn, where n represents any integer.

Hope you had fun on this tangent line adventure!

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

Let's start by finding the derivative of f(x).

First, apply the chain rule to the term sin 2x:
d/dx (sin 2x) = 2 * cos 2x

Then, apply the chain rule to the term -2 sin x:
d/dx (-2 sin x) = -2 * cos x

Now, add the derivatives of both terms to get the derivative of f(x):
f'(x) = 2 * cos 2x - 2 * cos x

To find the values of x where the derivative is equal to zero, we need to solve the equation f'(x) = 0.

Setting 2 * cos 2x - 2 * cos x = 0, we can simplify it further by factoring out a common factor of 2:
2 * (cos 2x - cos x) = 0

Since 2 is not equal to zero, we can divide both sides of the equation by 2:
cos 2x - cos x = 0

To solve this equation, we'll use the trigonometric identity:
cos 2x - cos x = 2 * sin^2(x) - 1 - cos x = 0

Rearranging this equation, we get:
2 * sin^2(x) - cos x - 1 = 0

Now, let's solve this quadratic equation for sin x:

To make this equation simpler, let's replace sin^2(x) with 1 - cos^2(x) using the Pythagorean Identity:
2 * (1 - cos^2(x)) - cos x - 1 = 0

Simplifying further:
2 - 2 * cos^2(x) - cos x - 1 = 0
-2 * cos^2(x) - cos x + 1 = 0

Now, we can solve this quadratic equation. Let's denote cos x as t:
-2t^2 - t + 1 = 0

To solve this equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values for a, b, and c, we get:
t = (-(-1) ± √((-1)^2 - 4 * (-2) * 1)) / (2 * (-2))
t = (1 ± √(1 + 8)) / (-4)
t = (1 ± √9) / (-4)
t = (1 ± 3) / (-4)

This gives us two possible values for t:
t1 = (1 + 3) / (-4) = 4 / (-4) = -1
t2 = (1 - 3) / (-4) = -2 / (-4) = 1/2

Since t = cos x, we can solve for x:
For t1 = -1, cos x = -1
This occurs when x = pi + 2n*pi, where n is an integer.

For t2 = 1/2, cos x = 1/2
This occurs when x = pi/3 + 2n*pi, where n is an integer.

Therefore, the x-coordinates of the points on the curve where the tangent line is horizontal are:
x = pi + 2n*pi, x = pi/3 + 2n*pi, where n is an integer.

In simpler terms:
x = pi(1 + 2n), x = pi/3(1 + 2n), where n is an integer.

To find the x-coordinates of all points on the curve where the tangent line is horizontal, we need to determine the values of x for which the derivative of the function f(x) is equal to zero.

Let's start by finding the derivative of f(x). The derivative of sin 2x is 2 cos 2x, and the derivative of -2 sin x is -2 cos x. Therefore, the derivative of f(x) is:

f'(x) = 2 cos 2x - 2 cos x

Now, set the derivative equal to zero and solve for x:

2 cos 2x - 2 cos x = 0

Factor out the common factor of 2:

2 (cos 2x - cos x) = 0

Now, we have two possibilities:

1. cos 2x - cos x = 0
2. 2 = 0 (this can be ignored since it leads to an invalid solution)

Let's solve the equation cos 2x - cos x = 0:

To simplify this, we can use the identity cos(2x) = 2cos^2(x) - 1:

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

Rearrange the equation:

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

Now, we have a quadratic equation. Let's solve it by factoring or using the quadratic formula. However, note that solving this equation may not result in nice, simple integer values for x.

Once you have found the solutions for x from the quadratic equation, list them as a comma-separated list. Remember to represent any integer solutions using "n."

Keep in mind that it is important to understand the process of finding solutions step by step, as solving the quadratic equation can be complex and may require the use of numerical methods if the solutions are not straightforward.