5.2 Solve the equation $\sqrt{x^{2}-4}+2 \sqrt{x^{2}-1}=\sqrt{7} \cdot x$.
5.2 It reminds one of Ptolemy's theorem. In a circle with diameter \(AB = x > 2\), take points \(C\) and \(D\) on the two semicircles such that \(AC = 1\), \(AD = 2\), then \(BD = \sqrt{x^2 - 4}\), \(BC = \sqrt{x^2 - 1}\). Thus, it can be deduced that \(CD = \sqrt{7}\), \(\angle CAD = 120^\circ\), \(x = \frac{2}{3} \sqrt{21}\).