Official solution
\[({\sin ^2}2x){{dy} \over {dx}} + (8{\sin ^2}2x + 2\sin 4x)y\]
\[ = 2{e^{ - 4x}}(2\sin 2x + \cos 2x)\]
\[{{dy} \over {dx}} + (8 + 4\cot 2x)y = 2{e^{ - 4x}}\left( {{{2\sin 2x + \cos 2x} \over {{{\sin }^2}2x}}} \right)\]
Integrating factor
\[(I.F.) = {e^{\int {(8 + 4\cot 2x)dx} }}\]
\[ = {e^{8x + 2\ln \sin 2x}}\]
Solution of differential equation
\[y.\,{e^{8x + 2\ln \sin 2x}}\]
\[ = \int {2{e^{(4x + 2\ln \sin 2x)}}{{(2\sin 2x + \cos 2x)} \over {{{\sin }^2}2x}}dx} \]
\[ = 2\int {{e^{4x}}(2\sin 2x + \cos 2x)dx} \]
\[y.\,{e^{8x + 2\ln \sin 2x}} = {e^{4x}}\sin 2x + c\]
\[y\left( {{\pi \over 4}} \right) = {e^{ - \pi }}\]
\[{e^{ - \pi }}\,.\,{e^{2\pi }} = {e^\pi } + c \Rightarrow c = 0\]
\[y\left( {{\pi \over 6}} \right) = {{{e^{{{2\pi } \over 3}}}{{\sqrt 3 } \over 2}} \over {{e^{\left( {{{4\pi } \over 3} + 2\ln {{\sqrt 3 } \over 2}} \right)}}}}\]
\[ = {e^{{{ - 2\pi } \over 3}}}\,.\,{2 \over {\sqrt 3 }}\]