Given that \(x=1+2\cos\theta\), prove that \(x^3-3x^2+1= 2\cos 3\theta-1\).

Using this result, or otherwise, find each of the three roots of the equation \(x^3-3x^2+1=0\), giving your answers correct to 3 places of decimals.

- The values of \(x\) and \(y\) satisfy the equations
\[\begin{align*}
3\cos^2 x+ 2\cos^2 y&=4,\\
3\sin 2x-2\sin 2y&=0.
\end{align*}\]
Find the values of \(\cos 2x\) and \(\cos 2y\).

Hence find, to the nearest degree, four pairs of corresponding values of \(x\) and \(y\), all values lying between \(0^\circ\) and \(360^\circ\).