Find the points on the curve `y = (cosx-1)` in `[0,2pi]`, where the tangent is parallel to `X-`axis.

Explanation

To find the points on the curve \( y = \cos x - 1 \) in the interval \([0, 2\pi]\) where the tangent is parallel to the x-axis, we need to follow these steps:

### Step 1: Understand the condition for the tangent to be parallel to the x-axis
The tangent to the curve is parallel to the x-axis when the derivative of the function is equal to zero. Therefore, we need to find the derivative of the function and set it to zero.

### Step 2: Find the derivative of the function
The given function is:
\[ y = \cos x - 1 \]
Now, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = -\sin x \]

### Step 3: 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:
\[ -\sin x = 0 \]
This simplifies to:
\[ \sin x = 0 \]

### Step 4: Solve for \(x\)
The solutions for \(\sin x = 0\) in the interval \([0, 2\pi]\) are:
\[ x = 0, \pi, 2\pi \]

### Step 5: Find the corresponding \(y\) values
Now, we will find the corresponding \(y\) values for each \(x\):
1. For \(x = 0\):
\[ y = \cos(0) - 1 = 1 - 1 = 0 \]
2. For \(x = \pi\):
\[ y = \cos(\pi) - 1 = -1 - 1 = -2 \]
3. For \(x = 2\pi\):
\[ y = \cos(2\pi) - 1 = 1 - 1 = 0 \]

### Step 6: List the points
The points on the curve where the tangent is parallel to the x-axis are:
- At \(x = 0\), the point is \((0, 0)\)
- At \(x = \pi\), the point is \((\pi, -2)\)
- At \(x = 2\pi\), the point is \((2\pi, 0)\)

Thus, the final answer is:
- The points are \((0, 0)\), \((\pi, -2)\), and \((2\pi, 0)\).