$y''+y'+y=\sin^2x$: particular solution? The problem I am trying to solve is finding the particular solution of the equation:

$$y''+y'+y=\sin^2x$$ 

I don't know what format the particular solution has. Once I know that, I can probably solve the problem with little difficulty. 
I haven't seen any examples in my textbook or else where with a exponential trig function on the right side. 
Using the format $y=Asin(x)+Bsin(x)$ (and therefore $y'=Acos(x)-Bsin(x)$, $y''=-Asin(x)-Bcos(x)$) and substituting these values for y and its derivatives doesn't give me any value with a $sin^2x$ in it.
What format does the specific solution have? How is this sort of equation supposed to be solved?
The solution of the characteristic equation $r^2+r+1=0$ is $(r+0.5)^2+0.75=0$ is $r=0.866i-0.5$; then $y_c=e^{0.5x}(Acos0.866x+Bsin0.866x)$ , but I don't know if this is is useful or at all how this is to be applied.

A particular solution is of the form
$$
y_p=A\cos(2x)+B\sin(2x)+C,
$$
for suitable values of $A,B,C$. Plugging the above expression in the ODE you obtain that
$$
y_p=\frac{1}{2}-\frac{1}{4}\sin (2x)
$$
