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

Question

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

${3 \over 8}$