If $x=2 \sin \theta-\sin 2 \theta$ and $y=2 \cos \theta-\cos 2 \theta, \theta \in[0,2 \pi]$, then $(\frac{d^2 y}{d x^2}) _{\theta=\pi}$ is : [JEE Main 2020, 9 Jan (Shift 2)]

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If \(x=2 \sin \theta-\sin 2 \theta\) and \(y=2 \cos \theta-\cos 2 \theta, \theta \in[0,2 \pi]\), then \((\frac{d^2 y}{d …

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