Explanation

(b)
Given that, $3\sin^{2}{A + 2\sin^{2}{B = 1}}$ …(i)
and $3\sin{2A - 2\sin{2B = 0}}$ …(ii)
From Eq. (i)
$3\left( \ \frac{1 - \cos{2A}}{2} \right) + 2\left( \ \frac{1 - \cos{2B}}{2} \right) = 1$
$\Rightarrow \ 3\cos{2A + 2\cos{2B = 3}}$ …(iii)
$\Rightarrow \ 3\cos{2A} = 3 - 2\cos{2B}$
$\Rightarrow \ 9\cos^{2}{2A = 9 + 4\cos^{2}{2B - 12\cos{2B}}}$
$\Rightarrow \ 9(1 - \sin^{2}{2A) = 9 + 4\cos^{2}{2B - 12\cos{2B}}}$
$\Rightarrow \ 9 - 4\sin^{2}{2B = 9 + 4\cos^{2}{2B - 12\ cos2B}}$ [from Eq. (ii)]
$\Rightarrow \ - 4(1 - \cos^{2}{2B\ ) = 4\cos^{2}{2B - 12\cos{2B}}}$
$\Rightarrow \ - 4 = - 12\cos{2B}$
$\Rightarrow \ \cos{2B = \frac{1}{3}}$
Now, from Eq. (iii)
$\cos{2A = \frac{7}{9}\ \Rightarrow \ 2\cos^{2}{A - 1 = \frac{7}{9}}}$
$\Rightarrow \ \cos{A = \frac{2\sqrt{2}}{3}}$
$\therefore\ A + 2B = \cos^{- 1}{\frac{2\sqrt{2}}{3} + \cos^{- 1}\frac{1}{3}}$
$= \cos^{- 1}\ \left( \frac{2\sqrt{2}}{3} \bullet \frac{1}{3} - \sqrt{1 - \frac{8}{9}}\ \sqrt{1 - \frac{1}{9}} \right)$
$= \cos^{- 1}\left( \frac{2\sqrt{2}}{9} - \frac{2\sqrt{2}}{9} \right)$
$= \cos^{- 1}{(0) = \frac{\pi}{2}}$