Original Function: -cos(3x)

Derivative Function: 3 sin(3x)
The period for the function and its derivative is 2π/3 radians. Over one period, when is the derivative function equal to its original function? That is, when is f(x) = f '(x) over 0≤x≤2π/3? Round your final answer to three decimal places.

Please help me with step by step solution algebraically

We want to find the values of x that satisfy the equation f(x) = f'(x), where f(x) = -cos(3x) and f'(x) = 3sin(3x).

So, we have -cos(3x) = 3sin(3x).

To solve this equation, we can rewrite it as -cos(3x) - 3sin(3x) = 0.

Next, let's use the trigonometric identity sin(A + B) = sin(A)cos(B) + cos(A)sin(B).

Using this identity, we can rewrite the equation as -cos(3x + π/2) = 0.

Now, solve for 3x + π/2 = π/2, since the cosine function equals 0 when the angle is π/2.

If we subtract π/2 from both sides, we get 3x = 0.

Divide both sides by 3, and we have x = 0.

So, one solution to the equation f(x) = f'(x) on the interval 0 ≤ x ≤ 2π/3 is x = 0.

Let's check if there are any more solutions.

We know that the period of the functions f(x) and f'(x) is 2π/3, which means that any value of x + k(2π/3), where k is an integer, will yield the same values for f(x) and f'(x).

Therefore, let's add 2π/3 to x and solve the equation again.

3(x + 2π/3) + π/2 = π/2

Expanding, we get 3x + 2π + π/2 = π/2

Subtract 2π from both sides, and we have 3x + π/2 = -3π/2.

Subtract π/2 from both sides, and we get 3x = -4π/2.

Simplify, and we have 3x = -2π.

Divide both sides by 3, and x = -2π/3.

So, another solution on the interval 0 ≤ x ≤ 2π/3 is x = -2π/3.

However, x = -2π/3 is not within the given interval. We need to find a solution within the interval 0 ≤ x ≤ 2π/3.

Subtract 2π/3 from x, and we get x = -2π/3 + 2π/3 = 0.

Therefore, the only solution that satisfies f(x) = f'(x) on the interval 0 ≤ x ≤ 2π/3 is x = 0.

Rounding x = 0 to three decimal places, we have x ≈ 0.000.

To find the points where the derivative function is equal to the original function, we can set them equal to each other and solve for x.

Given function: f(x) = -cos(3x)
Derivative function: f'(x) = 3sin(3x)

Setting them equal to each other:

-f(x) = f'(x)
-cos(3x) = 3sin(3x)

Next, we can use trigonometric identities to simplify the equation. Divide both sides by 3sin(3x):

-cos(3x) / 3sin(3x) = 1

Now, we can use the trigonometric identity cot(theta) = cos(theta) / sin(theta):

-cot(3x) = 1

Taking the inverse cotangent (or arccot) of both sides:

arccot(-cot(3x)) = arccot(1)

Since arccot(theta) can take on multiple values, we'll consider the principal value, which is usually in the range (-π/2, π/2).

arccot(-cot(3x)) = π/4

Now we need to solve for x. To do that, we can rewrite the equation:

3x = arccot(1) + πn, where n is an integer.

Dividing both sides by 3:

x = (arccot(1) + πn) / 3

To find the values of x within the interval 0 ≤ x ≤ 2π/3, we substitute n = 0, 1, and 2 into the equation.

For n = 0:
x = (arccot(1) + 0) / 3 = arccot(1) / 3

For n = 1:
x = (arccot(1) + π) / 3

For n = 2:
x = (arccot(1) + 2π) / 3

To get the final answer, we substitute arccot(1) = π/4 into the equations and round the x-values to three decimal places.

For n = 0:
x = (π/4) / 3 ≈ 0.262

For n = 1:
x = (π/4 + π) / 3 ≈ 1.049

For n = 2:
x = (π/4 + 2π) / 3 ≈ 1.836

Therefore, over the interval 0 ≤ x ≤ 2π/3, the derivative function is equal to the original function at approximately x = 0.262, 1.049, and 1.836 (rounded to three decimal places).