If A and B are acute angles satisfying $3 {\sin }^{2}A + 2 {\sin }^{2}B=1$ and 3 sin2A = 2sin2B then cos(A + 2B) =

$$\begin{gathered} 3{\sin }^2\mathrm{A}+2{\sin }^2\mathrm{B}=1 \\ 3{\sin }^2\mathrm{A}=1-2{\sin }^2\mathrm{B} \\ 3{\sin }^2\mathrm{A}=\cos 2\mathrm{B} \\ 3\sin 2\mathrm{A}=2\sin 2\mathrm{B} \\ \sin 2\mathrm{B}=\frac{3}{2}\sin 2\mathrm{A} \\ \cos (\mathrm{A}+2\mathrm{B})=\cos \mathrm{A} \cos 2\mathrm{B}-\sin \mathrm{A} \sin 2\mathrm{B} \\ =\mathrm{cosA} 3{\sin }^2\mathrm{A}-\sin \mathrm{A}\frac{3}{2} \sin 2\mathrm{A} \\ =3{\sin }^2\mathrm{A} \cos \mathrm{A}-\left(\frac{3}{2}\right) \sin \mathrm{A} \cos \mathrm{A} \sin \mathrm{A} \\ =3{\sin }^2 \mathrm{A} \cos \mathrm{A}-3{\sin }^2\mathrm{A} \mathrm{cosA} \\ =0 \end{gathered}$$