If x=2 sin θ- sin 2 θ and y=2 cos θ- cos 2 θ, θ ∈[0,2 π], then (d^2 y)/(d x^2) at θ=π is:

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