Answer: \(\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=\frac{\pi}{6}}=-4\), and \(P\left(\frac{\pi}{6}-4\sqrt3,0\right)\).

Using the product rule on \(y=4\sin2x\cos2x\),

\(\frac{\mathrm{d}y}{\mathrm{d}x}=4\left(2\cos2x\cos2x-2\sin2x\sin2x\right).\)

So

\(\frac{\mathrm{d}y}{\mathrm{d}x}=8(\cos^2 2x-\sin^2 2x).\)

At \(x=\frac{\pi}{6}\), \(2x=\frac{\pi}{3}\). Hence

\(\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=\frac{\pi}{6}}=8\left(\frac14-\frac34\right)=-4.\)

Now find the point on the curve:

\(y=4\sin\frac{\pi}{3}\cos\frac{\pi}{3}=4\cdot\frac{\sqrt3}{2}\cdot\frac12=\sqrt3.\)

The gradient of the normal is the negative reciprocal of \(-4\), so it is \(\frac14\).

The normal through \(\left(\frac{\pi}{6},\sqrt3\right)\) is

\(y-\sqrt3=\frac14\left(x-\frac{\pi}{6}\right).\)

At the \(x\)-axis, \(y=0\), so

\(-\sqrt3=\frac14\left(x-\frac{\pi}{6}\right).\)

Thus

\(x=\frac{\pi}{6}-4\sqrt3.\)

Therefore

\(P\left(\frac{\pi}{6}-4\sqrt3,0\right).\)