For example, with the point $\,\displaystyle{\left(-\frac{2}{\sqrt 3}\,,\,\frac{3}{2}\right)}\,$ , and with $\,t=\theta\,$ , for simplicity:
$2\cot t=x=-\frac{2}{\sqrt 3}\Longrightarrow \tan t=-\sqrt 3\Longrightarrow t=\frac{\pi}{3}+k\pi\,\,,\,k\in\Bbb Z$
$\frac{3}{2}=y=2\sin^2t\Longrightarrow \sin t=\pm\frac{\sqrt 3}{2}\Longrightarrow t=\frac{\pi}{3}+k\pi\,\,\,,\,\,k\in\Bbb Z$
Well, choose one of the infinite ammount of possible $\,t'$s above, say
$t=\frac{\pi}{3}\,\,(k=0\,)\Longrightarrow\,\left.\frac{dy}{dx}\right|_{t=\pi/3}=-2\sin^3\frac{\pi}{3}\cos\frac{\pi}{3}=-2\left(\frac{\sqrt 3}{2}\right)^3\frac{1}{2}=-\frac{3\sqrt 3}{8}$
so the tangent line to the curve at this point is
$y-\frac{3}{2}=-\frac{3\sqrt 3}{8}\left(x+\frac{2}{\sqrt 3}\right)$