(i) To find the \(x\)-coordinates of \(A\) and \(B\), set the equations equal: \(2x^5 + 3x^3 = 2x\).

Rearrange to: \(2x^5 + 3x^3 - 2x = 0\).

Factor out \(x\): \(x(2x^4 + 3x^2 - 2) = 0\).

Thus, \(2x^4 + 3x^2 - 2 = 0\) is satisfied by the \(x\)-coordinates of \(A\) and \(B\).

(ii) Solve \(2x^4 + 3x^2 - 2 = 0\) by substituting \(y = x^2\), giving \(2y^2 + 3y - 2 = 0\).

Factorize: \((y + 2)(2y - 1) = 0\).

Thus, \(y = -2\) or \(y = \frac{1}{2}\).

Since \(y = x^2\), \(x^2 = \frac{1}{2}\) (ignore \(y = -2\) as \(x^2\) cannot be negative).

So, \(x = \pm \frac{1}{\sqrt{2}}\).

For \(x = \frac{1}{\sqrt{2}}\), \(y = 2x = \frac{2}{\sqrt{2}}\).

For \(x = -\frac{1}{\sqrt{2}}\), \(y = 2x = -\frac{2}{\sqrt{2}}\).

Thus, the coordinates are \(\left( \frac{1}{\sqrt{2}}, \frac{2}{\sqrt{2}} \right)\) and \(\left( -\frac{1}{\sqrt{2}}, -\frac{2}{\sqrt{2}} \right)\).