Official solution
\[2\cos \left( {{{x + y} \over 2}} \right)\cos \left( {{{x - y} \over 2}} \right) - \left[ {2{{\cos }^2}\left( {{{x + y} \over 2}} \right) - 1} \right] = {3 \over 2}\]
\[2\cos \left( {{{x + y} \over 2}} \right)\left[ {\cos \left( {{{x - y} \over 2}} \right) - \cos \left( {{{x + y} \over 2}} \right)} \right] = {1 \over 2}\]
\[2\cos \left( {{{x + y} \over 2}} \right)\left[ {2\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right)} \right] = {1 \over 2}\]
\[\cos \left( {{{x + y} \over 2}} \right).\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right) = {1 \over 8}\]
Possible when \[{x \over 2} = 30^\circ \] \& \[{y \over 2} = 30^\circ \]
\[x = y = 60^\circ \]
\[\sin x + \cos y = {{\sqrt 3 } \over 2} + {1 \over 2} = {{\sqrt 3 + 1} \over 2}\]