Yes, you computed the derivative correctly (assuming you meant to write $(\sin x)^2$ or $\sin^2 x$, not $\sin x^2$).
You need to find the points where the derivative is zero. First note that since $\sin$ is bounded in absolute value by 1, the denominator in your expression for the derivative is never zero. So, the zeroes of the derivative are precisely the zeroes of the numerator of your expression. You thus have to solve the equation
$$
(-2\sin x -\sin^2 x)-\cos^2 x=0;
$$
a task that is made tractable upon using the Pythagorean identity $\sin^2 x+\cos^2 x=1$. Utilizing this (and a bit of algebra) produces the equivalent equation
$$
-2\sin x-1=0.
$$
Can you take it from here?
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.
Best Answer
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 $x$