(b) Since $\alpha$ is a root of $25{\cos ^2}\theta + 5\cos \theta – 12 = 0$

$\therefore$ $25{\cos ^2}\alpha + 5\cos \alpha – 12 = 0$

==> $\cos \alpha = \frac{{ – 5 \pm \sqrt {25 + 1200} }}{{50}}$ $ = \frac{{ – 5 \pm 35}}{{50}}$

==> $\cos \alpha = – 4/5$     $[ \because \pi /2 < \alpha  < \pi  \Rightarrow \cos \alpha  < 0]$

$\therefore $$\sin 2\alpha = 2\sin \alpha \cos \alpha = – 24/25$.