Answer: \(\displaystyle P\left(\frac{4\pi}{3}-\frac{\sqrt3}{16},0\right)\).

For a trigonometric graph, identify the amplitude, midline and period first, then use these to locate the key points of the sketch.

The point on the curve has

\(x=\frac{4\pi}{3}.\)

So

\(y=\cos\left(\frac14\cdot\frac{4\pi}{3}\right) =\cos\frac{\pi}{3} =\frac12.\)

Thus the point on the curve is

\(\left(\frac{4\pi}{3},\frac12\right).\)

Differentiate \(y=\cos\frac{x}{4}\):

\(\frac{dy}{dx}=-\frac14\sin\frac{x}{4}.\)

At \(x=\frac{4\pi}{3}\),

\(\frac{dy}{dx}=-\frac14\sin\frac{\pi}{3} =-\frac{\sqrt3}{8}.\)

This is the gradient of the tangent. Therefore the gradient of the normal is

\(\frac{8}{\sqrt3}.\)

The equation of the normal is

\(y-\frac12=\frac{8}{\sqrt3}\left(x-\frac{4\pi}{3}\right).\)

At the \(x\)-axis, \(y=0\). Substitute this:

\(-\frac12=\frac{8}{\sqrt3}\left(x-\frac{4\pi}{3}\right).\)

So

\(x-\frac{4\pi}{3}=-\frac{\sqrt3}{16}.\)

Hence

\(x=\frac{4\pi}{3}-\frac{\sqrt3}{16}.\)

Therefore

\(P\left(\frac{4\pi}{3}-\frac{\sqrt3}{16},0\right).\)