parametric equations of a curve The parametric equations of a curve are $$x=2\theta-\sin 2\theta$$ $$y=2-\cos 2\theta$$
The question asks that ''For the part of the curve where $0<\theta<2\pi$, find the coordinates of the points where the tangent is parallel to the x-axis''. How can I solve this? 
 A: $\frac{dy}{d\theta} = 2\sin2\theta$.
$\frac{dx}{d\theta} = 2 - 2\cos2\theta$
So
$\frac{dy}{dx} = \frac{dy}{d\theta}/\frac{dx}{d\theta} = \frac{2\sin2\theta}{2 - 2\cos2\theta}$.
This simplifies to $\cot\theta$, and since we want slope to be zero, we get $\theta = \frac{\pi}{2}$ or $\theta = \frac{3\pi}{2}$ (given the values $\theta$ can take)
At $\theta = \frac{\pi}{2}$, $x = \pi$ and $y = 3$.
At $\theta = \frac{3\pi}{2}$, $x = 3\pi$ and $y = 3$.
So the points at which the tangent is parallel to the x-axis is $\color{blue}{(\pi,3), (3\pi,3)}$.
A: It is clear that when $\theta\in \mathbb R$ the only extrema of the curve are $y=2-1=1$ (minimum) and $y=2-(-1)=3$ (maximum). It is clear too that in the interval $0<\theta<2\pi$ there are just three values of $\theta$ to be considered: $\frac{\pi}{2}, \pi, \frac{3\pi}{2}$, $\pi$ giving minimum and the two other giving maximum.
Taking into account the analytical expression of  $x$ and $y$, the values of $\theta$ giving points for which the tangent to the curve is parallel to the x-axis correspond to the values for which the derivative $\frac{dy}{dx}$ is well defined and equal to zero. 
We have $$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$ 
Hence$$\frac{dy}{dx}=\frac{2sin 2\theta}{2-2cos 2\theta}=0\Rightarrow sin 2\theta=0$$
It is immediate that  $\frac{\pi}{2}$ and $ \frac{3\pi}{2}$ give two solutions but 
for $\pi$ the derivative take the form $\frac{0}{0}$ (it is easy to prove that it tends to $\infty$ when $\theta\to \pi$ so there are no well defined tangent but a limit vertical position of it). There are actually infinitely many tangents at point $(2\pi, 1)$ which is however a minimum indeed.
The answer is $\boxed{(x,y)=(\pi,3),(3\pi,3)}$. The period is equal to $\pi$.
NOTE.- Just to illustrate how the parametric equations can be useful, the Cartesian equation of the curve is $$x^2+y^2-2y+3-2x(arc\space cos(y-2))+[arc\space cos(y-2)]^2=0$$ 
