Bài giải:

$2.{{\sin }^{2}}\alpha +3.{{\cos }^{2}}\alpha =\frac{9}{4}\Leftrightarrow 2.({{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha )+{{\cos }^{2}}\alpha =\frac{9}{4}$

$\Leftrightarrow 2.1+{{\cos }^{2}}\alpha =\frac{9}{4}\Leftrightarrow {{\cos }^{2}}\alpha =\frac{9}{4}-2=\frac{1}{4}$

Vì ${{0}^{o}}<\alpha <{{90}^{o}}$nên $\cos \alpha >0$

$\Rightarrow \cos \alpha =\frac{1}{2}\Rightarrow \alpha ={{60}^{o}}$