Answer: (i) A suitable sketch is one loop of the rose curve with maximum radius at \(\theta=0\), so it passes through \((a,0)\) and returns to the origin at \(\theta=\pm \frac{\pi}{6}\).

(ii) The enclosed area is \(\frac{\pi a^{2}}{12}\).

(iii) A cartesian equation of \(C\) is \(\left(x^{2}+y^{2}\right)^{2}=a x\left(x^{2}-3y^{2}\right)\).

(i) Since \(r=a\cos 3\theta\) and \(-\frac{\pi}{6}\leqslant \theta \leqslant \frac{\pi}{6}\), we have \(\cos 3\theta\geqslant 0\), so \(r\geqslant 0\) throughout this interval. As \(\theta\) runs from \(-\frac{\pi}{6}\) to \(\frac{\pi}{6}\), the curve forms one loop.

At \(\theta=0\), \(r=a\), so the curve reaches the point \((a,0)\). At \(\theta=\pm \frac{\pi}{6}\), \(3\theta=\pm \frac{\pi}{2}\), so \(r=0\), meaning the curve starts and ends at the origin. The loop is symmetric about the polar axis.

(ii) The area enclosed by a polar curve is

\(A=\frac{1}{2}\int r^{2}\,d\theta\).

Here \(r=a\cos 3\theta\), so

\(A=\frac{1}{2}\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} a^{2}\cos^{2}3\theta\,d\theta\).

Use \(\cos^{2}x=\frac{1}{2}(1+\cos 2x)\):

\(A=\frac{a^{2}}{2}\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\frac{1}{2}(1+\cos 6\theta)\,d\theta\)

\(=\frac{a^{2}}{4}\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}(1+\cos 6\theta)\,d\theta\).

Integrating gives

\(A=\frac{a^{2}}{4}\left[\theta+\frac{1}{6}\sin 6\theta\right]_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\).

Now \(\sin(\pi)=\sin(-\pi)=0\), so

\(A=\frac{a^{2}}{4}\left(\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)\right)=\frac{a^{2}}{4}\cdot\frac{\pi}{3}=\frac{\pi a^{2}}{12}.\)

(iii) Starting from \(r=a\cos 3\theta\) and using \(\cos 3\theta=4\cos^{3}\theta-3\cos\theta\),

\(r=a\cos\theta\left(4\cos^{2}\theta-3\right).\)

Now \(x=r\cos\theta\), so \(\cos\theta=\frac{x}{r}\), and also \(r^{2}=x^{2}+y^{2}\).

Substitute into the polar equation:

\(r=a\left(\frac{x}{r}\right)\left(4\left(\frac{x}{r}\right)^{2}-3\right).\)

Multiply through by \(r^{3}\):

\(r^{4}=ax\left(4x^{2}-3r^{2}\right).\)

Since \(r^{2}=x^{2}+y^{2}\), this becomes

\(\left(x^{2}+y^{2}\right)^{2}=ax\left(4x^{2}-3(x^{2}+y^{2})\right).\)

So

\(\left(x^{2}+y^{2}\right)^{2}=ax\left(x^{2}-3y^{2}\right).\)