At what points on the following curve, is the tangent parallel to x-axis ?
`y=cosx-1" on "[0,2pi]`

Answer

To find the points on the curve \( y = \cos x - 1 \) 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 its slope is zero. The slope of the curve at any point is given by the derivative \( \frac{dy}{dx} \).

### Step 2: Differentiate the function
We differentiate the function \( y = \cos x - 1 \):
\[ \frac{dy}{dx} = -\sin x \]

### Step 3: Set the derivative equal to zero
To find where the slope is zero, we set the derivative equal to zero:
\[ -\sin x = 0 \]
This simplifies to:
\[ \sin x = 0 \]

### Step 4: Solve for \( x \)
The general solution for \( \sin x = 0 \) is:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]
Now, we need to find the values of \( x \) in the interval \( [0, 2\pi] \).

### Step 5: Determine the specific values of \( x \)
In the interval \( [0, 2\pi] \), the values of \( n \) can be:
- For \( n = 0 \): \( x = 0 \)
- For \( n = 1 \): \( x = \pi \)
- For \( n = 2 \): \( x = 2\pi \)

Thus, the points where the tangent is parallel to the x-axis are \( x = 0, \pi, 2\pi \).

### Step 6: Find the corresponding \( y \)-values
Now we substitute these \( x \)-values back into the original equation to find the corresponding \( y \)-values:
1. For \( x = 0 \):
\[ y = \cos(0) - 1 = 1 - 1 = 0 \quad \Rightarrow \quad (0, 0) \]
2. For \( x = \pi \):
\[ y = \cos(\pi) - 1 = -1 - 1 = -2 \quad \Rightarrow \quad (\pi, -2) \]
3. For \( x = 2\pi \):
\[ y = \cos(2\pi) - 1 = 1 - 1 = 0 \quad \Rightarrow \quad (2\pi, 0) \]

### Final Points
Thus, the points on the curve where the tangent is parallel to the x-axis are:
- \( (0, 0) \)
- \( (\pi, -2) \)
- \( (2\pi, 0) \)