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

$$\begin{gathered} \mathrm{x}=2 \mathrm{sin\theta }-\sin 2\mathrm{\theta }, \mathrm{y}=2\cos \mathrm{\theta }-\cos 2\mathrm{\theta } \\ \frac{\mathrm{dx}}{\mathrm{d\theta }}=2\cos \mathrm{\theta }-2\cos 2\mathrm{\theta } \\ \frac{\mathrm{dy}}{\mathrm{d\theta }}=-2\mathrm{sin\theta }+2\sin 2\mathrm{\theta } \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\displaystyle \frac{\mathrm{dy}}{\mathrm{d\theta }}}}{{\displaystyle \frac{\mathrm{dx}}{\mathrm{d\theta }}}}=\frac{2(\sin 2\mathrm{\theta }-\mathrm{sin\theta })}{2(\mathrm{cos\theta }-\cos 2\mathrm{\theta })} \\ =\frac{2\cos {\displaystyle \frac{3\mathrm{\theta }}{2}}\sin {\displaystyle \frac{\mathrm{\theta }}{2}}}{2\sin {\displaystyle \frac{3\mathrm{\theta }}{2}}\sin {\displaystyle \frac{\mathrm{\theta }}{2}}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\cot \frac{3\mathrm{\theta }}{2} \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{-3}{2}{\mathrm{cosec}}^2\frac{3\mathrm{\theta }}{2}\frac{\mathrm{d\theta }}{\mathrm{dx}} \\ =\frac{-3}{2}{\mathrm{cosec}}^2\frac{3\mathrm{\theta }}{2}.\frac{1}{2\mathrm{cos\theta }-2\cos 2\mathrm{\theta }} \\ \mathrm{\theta }=\mathrm{\pi }\Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{-3}{2}{(-1)}^2\frac{1}{-2-2} =\frac{-3}{2}\frac{1}{(-4)}=\frac{3}{8} \end{gathered}$$