The derivative of this curve can be computed using the implicit derivative method.

$$\frac{dx}{dt} = -\frac{1}{\sin^2{t}}\quad\text{and}\quad \frac{dy}{dt} = \frac{-2\sin(2t)\sin{t}-\cos(2t)\cos{t}}{\sin^2{t}}$$

$$\therefore\quad\frac{dy}{dx}=2\sin(2t)\sin{t}+\cos(2t)\cos{t}= 4\sin^2{t}\cos{t} + (1-2\sin^2{t})\cos{t}= (1+2\sin^2{t})\cos{t}$$

It follows that $$\begin{align*} \frac{d^2y}{dx^2} &= 4\sin{t}\cos^2{t} - (1+2\sin^2{t})\sin{t}\\ &= \sin{t}(4\cos^2{t} - 1 - 2\sin^2{t})\\ &= \sin{t}(4\cos^2{t}- 1 -2 + 2\cos^2{t})\\ &= 3\sin{t}(2\cos^2{t}-1)\\ &= -3\sin{t}\cos(2t)\end{align*}$$

Setting $\frac{d^2y}{dx^2}=0$ leads to $t_1=\frac{\pi}{4}$ and $t_2=\frac{3\pi}{4}$. Meanwhile,

• when $0 < t < \frac{\pi}{4}$, $\frac{d^2y}{dx^2} < 0$.
  
• when $\frac{\pi}{4} < t < \frac{3\pi}{4}$, $\frac{d^2y}{dx^2} > 0$.
• when $\frac{3\pi}{4} < t < \pi$, $\frac{d^2y}{dx^2} < 0$.

Therefore, both points are inflection points. Their corresponding coordinates are $\boxed{(\pm 1, 0)}$.