Answer
$c_1cos2x+c_2sin2x-\frac{1}{4}xcos2x$
Work Step by Step
$y''+4y=sin2x$ which has characteristic equation $t^2+4=0$ with solutions $t=\pm 2i$
Particular solution: $Axcos2x+Bxsin2x$ giving $A=-1/4$ and $B=0$
General solution: $c_1cos2x+c_2sin2x-\frac{1}{4}xcos2x$