$\text{ If }x=2\sin \theta -\sin 2\theta \text{ and }y=2\cos \theta -\cos 2\theta ,\theta \in [0,2\pi ],\text{ then }\frac{d^{2}y}{d{x}^2}at \theta =\pi \text{ is: }$

$\begin{array}{l}x=2\sin \theta -\sin 2\theta y=2\cos \theta -\cos 2\theta \\ \frac{dx}{d\theta }=2\cos \theta -2\cos 2\theta \frac{dy}{d\theta }=-2\sin \theta +2\sin 2\theta \\ \frac{dy}{dx}=\frac{-2\sin \theta +2\sin 2\theta }{2\cos \theta -2\cos 2\theta }=\frac{-\sin \theta +\sin 2\theta }{\cos \theta -\cos 2\theta }\\ \frac{d^{2}y}{d{x}^2}=\frac{(\cos \theta -\cos 2\theta )(-\cos \theta +2\cos 2\theta )-(\sin 2\theta -\sin \theta )(-\sin \theta +2\sin 2\theta )}{(\cos \theta -\cos 2\theta {)}^2}\times \frac{d\theta }{dx}\\ \frac{d^{2}y}{d{x}^2(\theta =\pi )}=\frac{(-1-1)(1+2)-\left(0\right)}{(-1-1{)}^2}\times \left(\frac{-1}{4}\right)=\frac{3}{8}\end{array}$