Using Rolle's theorem find the point in `( 0, 2 pi)` on the curve `y = cosx - 1,` where tangent is parallel to x axis.

Explanation

To solve the problem using Rolle's theorem, we will follow these steps:

### Step 1: Define the function
We are given the function:
\[ f(x) = \cos x - 1 \]

### Step 2: Check the conditions of Rolle's Theorem
Rolle's theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one point \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

- **Continuity**: The function \( f(x) = \cos x - 1 \) is continuous everywhere since cosine is continuous.
- **Differentiability**: The function is differentiable everywhere as well.
- **Endpoints**: Now we check \( f(0) \) and \( f(2\pi) \):
\[ f(0) = \cos(0) - 1 = 1 - 1 = 0 \]
\[ f(2\pi) = \cos(2\pi) - 1 = 1 - 1 = 0 \]
Since \( f(0) = f(2\pi) \), we can apply Rolle's theorem.

### Step 3: Differentiate the function
Next, we find the derivative of \( f(x) \):
\[ f'(x) = -\sin x \]

### Step 4: Set the derivative equal to zero
To find the points where the tangent is parallel to the x-axis, we set the derivative equal to zero:
\[ -\sin x = 0 \implies \sin x = 0 \]

### Step 5: Solve for \( x \)
The solutions to \( \sin x = 0 \) are:
\[ x = n\pi \quad \text{for integers } n \]
In the interval \( (0, 2\pi) \), the valid solution is:
\[ x = \pi \]

### Step 6: Find the corresponding \( y \)-coordinate
Now we substitute \( x = \pi \) back into the original function to find the \( y \)-coordinate:
\[ f(\pi) = \cos(\pi) - 1 = -1 - 1 = -2 \]

### Final Result
Thus, the point in \( (0, 2\pi) \) on the curve where the tangent is parallel to the x-axis is:
\[ (\pi, -2) \]

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