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

The equation of the curve is $y = \cos x - 1$.

Now, we have to find a point on the curve in $[0,2\pi ]$,

where the tangent is parallel to $X$-axis i.e., the tangent to the curve at $x = c$ has a slope o, where $c \in (0,2\pi )$.

Let us apply Rolle's theorem to get the point.

(i) $y = \cos x - 1$ is a continuous function in $[0,2\pi ]$.
[since it is a combination of cosine function and a constant function]

(ii) $y' = - \sin x,$ which exists in $(0,2\pi )$.

Hence, $y$ is differentiable in $(0,2\pi )$.

(iii) $y(0) = \cos 0 - 1 = 0$ and $y(2\pi ) = \cos 2\pi - 1 = 0$,
therefore,$y(0) = y(2\pi )$

Since, conditions of Rolle's theorem are satisfied.

Hence, there exists a real number $c$ such that
$f(c) = 0$

$\Rightarrow$ $- \sin c = 0$

$\Rightarrow$ $c = \pi$ or 0, where $\pi \in (0,2\pi )$

$\Rightarrow$ $x = \pi$

therefore,$y = \cos \pi - 1 = - 2$

Hence, the required point on the curve, where the tangent drawn is parallel to the $X$-axis is $(\pi , - 2)$.