If you have a parametric equation $x=x(t)$ and $y=y(t)$, then
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$
In your case, you have $x(\theta)=r\theta-r\sin\theta$ and $y(\theta)=r-r\cos\theta$. Therefore, $\dfrac{dx}{d\theta} = \ldots$ and $\dfrac{dy}{d\theta} = \ldots$, and hence
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \ldots$$
Then you'll want to compute $\left.\dfrac{dy}{dx}\right|_{\theta=\pi/6}$ and $\left.\dfrac{dy}{dx}\right|_{\theta=5\pi/4}$ to get the slopes of the two tangent lines. Once you have those, you'll be able to compute the tangent line equations like usual using point-slope form; in particular, your two tangent line equations will be
$$y=\left(\left.\dfrac{dy}{dx}\right|_{\theta=\pi/6}\right)(x-x(\pi/6)) + y(\pi/6)$$
and
$$y=\left(\left.\dfrac{dy}{dx}\right|_{\theta=5\pi/4}\right)(x-x(5\pi/4)) + y(5\pi/4)$$
To see where the tangent line is horizontal, you'll want to solve $\dfrac{dy}{d\theta}=0$.
To see where the tangent line is vertical, you'll want to solve $\dfrac{dx}{d\theta}=0$.
To see if the cycloid is concave up or down, you'll need to compute the second derivative:
$$\frac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\frac{dy}{dx}\right) = \dfrac{\dfrac{d}{d\theta}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{d\theta}}$$
and see where it's positive and/or negative.
I hope this provides you with enough information to at least start the problem.
The gradient$(m)$ of the tangent line $=f'(x)$
The tangent line will be horizontal of $y=f(x)$ if $f'(x)=0$ and will be vertical if $\displaystyle f'(x)=\infty\implies \frac1{f'(x)}=0$
Now, here $\displaystyle y=\sin2x+2\sin x\implies \frac{dy}{dx}=2\cos2x+2\cos x$
So, horizontal tangent, we need $\cos2x+\cos x=0\iff \cos2x=-\cos x=\cos(\pi-x)$
$\displaystyle\implies 2x=2n\pi\pm(\pi-x)$ where $n$ is any integer
Taking the '+' sign, $\displaystyle\implies 2x=2n\pi+(\pi-x)\implies x=\frac{(2n+1)\pi}3$
Now $0\le x<2\pi\implies 0\le \frac{(2n+1)\pi}3<2\pi\iff 0\le 2n+1<6\iff0\le n\le2$
Similarly, for the '-' sign
Now as $\displaystyle|\cos y|\le1$ for real $\displaystyle y, \frac{dy}{dx}=2\cos2x+2\cos x$ will be finite, hence no vertical tangent line
In fact, we can find the range of $\displaystyle 2\cos2x+2\cos x$ for real $x$
Best Answer
Hint 1. A line is horizontal when its slope is $0$.
Hint 2. The slope of the tangent line to $y=f(x)$ at $a$ is $f'(a)$.