Answer
The points with a horizontal tangent line on the cycloid are
$c((2n-1) \pi)=((2n-1) \pi,2)$ for $n=0,\pm 1,\pm 2,\pm 3,...$.
Work Step by Step
The cycloid generated by the unit circle has parametric equations:
$x(t)=t-\sin t$, $y(t)=1-\cos t$.
Using Eq. (8) the slope of the tangent line is
$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{\sin t}{1- \cos t}$.
The tangent line is horizontal if $\frac{dy}{dx}=0$. So, we solve the equation $\frac{\sin t}{1-\cos t}=0$.
Notice that $\frac{dy}{dx}$ is infinite if $\cos t=1$. So, the required conditions are $\sin t=0$ and $\cos t \neq 1$.
The solutions are $t=(2n-1) \pi$, where $n=0,\pm 1,\pm 2,\pm 3,...$. The corresponding points are
$c((2 n-1)\pi)=((2 n-1)\pi-\sin (2 n-1)\pi,1-\cos (2 n-1)\pi)$
for $n=0,\pm 1,\pm 2,\pm 3,...$.
Since $\sin (2n-1) \pi=0$ and $\cos (2n-1) \pi=-1$ for $n=0,\pm 1,\pm 2,\pm 3,...$, the points with a horizontal tangent line on the cycloid are
$c((2n-1) \pi)=((2n-1) \pi,2)$
for $n=0,\pm 1,\pm 2,\pm 3,...$.
