A curve is such that \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=8\sin 4x-4\cos 2x\).
The curve passes through the point \(P\left(\pi,-\frac{1}{6}+\pi\right)\) and has a gradient of \(-1\) at \(P\).
Find the equation of the curve.[7]
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