Show that \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac16\tan^3 2x\right)=\sec^4 2x-\sec^2 2x\).[3]
For a particular curve where \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=9\cos3x-5\sin x\), the curve passes through the point \(P\left(\dfrac{3\pi}{2},\dfrac{15\pi}{2}\right)\) and the gradient of the curve at \(P\) is 8. Find the equation of the curve.[6]
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