Q.Find the points on the curve $y = (\cos x - 1)$ in $[0, 2\pi]$, where the tangent is parallel to the $x$-axis.

The key idea is that a tangent parallel to the $x$-axis means the slope $\frac{dy}{dx} = 0$.

Step 1: Differentiate $y = \cos x - 1$ with respect to $x$:

Step 2: Set the derivative equal to zero:

Step 3: Solve $\sin x = 0$ in the interval $[0, 2\pi]$. The solutions are:

$x = 0,\ \pi,\ 2\pi.$ …

The key idea here is simple: a line parallel to the $x$-axis has slope zero. So we are really asking: at what points on this curve does the derivative vanish? That is the entire conceptual backbone — no parametric tricks needed here, just a straightforward first-derivative analysis.

Let’s walk through it.

1. Understand the condition.

   A tangent line parallel to the $x$-axis means its slope is $0$. The slope of the tangent to $y = f(x)$ at any point is $\frac{dy}{dx}$. So we set $\frac{dy}{dx} = 0$ and solve for $x$.

2. Differentiate the given function.

   $y = \cos x - 1$

   $\frac{dy}{dx} = -\sin x$

   (Derivative of $\cos x$ is $-\sin x$, and the constant $-1$ differentiates to $0$.)

3. Set the derivative to zero.

   $-\sin x = 0 \implies \sin x = 0$

4. Solve $\sin x = 0$ in $[0, 2\pi]$.

   The sine function is zero at integer multiples of $\pi$:

   $x = 0, \pi, 2\pi$

   All three lie in the closed interval $[0, 2\pi]$.

5. Find the corresponding $y$-coordinates.
   • At $x = 0$: $y = \cos 0 - 1 = 1 - 1 = 0$ …

Method: Finding Points Where the Tangent Is Parallel to the $x$-Axis

Use this method for any curve $y=f(x)$ where you must locate every point at which the tangent line is horizontal.

Steps

Step 1: Translate the geometric condition into an equation

A horizontal line has slope $0$, and the tangent's slope at any point on $y=f(x)$ is $\dfrac{dy}{dx}$. So the condition "tangent parallel to the $x$-axis" becomes:

Step 2: Differentiate $y=f(x)$

Apply the standard derivative rules (power, chain, trig, etc.) to find $\dfrac{dy}{dx}$ as a function of $x$.

Step 3: Solve $\dfrac{dy}{dx}=0$ within the given interval …

Common Mistakes

Mistake 1: Differentiating the constant $-1$ as if it contributed a term

Why it's wrong: $y=\cos x-1$ has $\dfrac{dy}{dx}=-\sin x$ — the $-1$ is a constant shift and differentiates to $0$, but a rushed student can mistakenly carry it through as an extra $-1$ in the derivative, writing $\dfrac{dy}{dx}=-\sin x-1$, which changes where the slope actually equals zero. Correct approach: remember a constant term vanishes completely under differentiation; only the $\cos x$ part contributes.

Mistake 2: Dismissing $x=0$ and $x=2\pi$ because they're endpoints …