The differential equation is y''+ 2y'+ y= A sin(x). The associated homogeneous equation is y''+ 2y'+ y= 0 which has characteristic equation \(\displaystyle r^2+ 2r+ 1= (r+ 1)^2= 0\). That has the single solution r= -1 so the general solution to the homogeneous equation is \(\displaystyle y(x)= C_1e^{-x}+ C_2xe^{-x}\).
You appear to have done that part correctly. Now, knowing that the derivative of sin(x) is cos(x) and that the derivative of cos(x) is -sin(x), I would look for a solution to the entire equation \(\displaystyle y(x)= A sin(x)+ B cos(x)\) for constants A and B.
y'(x)= A cos(x)- B sin(x) and y''(x)= -A sin(x)- B cos(x) so that y''+ 2y'+ y= -A sin(x)- B cos(x)+ 2A cos(x)- 2B sin(x)+ A sin(x)+ B cos(x)= 2B sin(x)+ 2A cos(x)= sin(x). Since this is to be true for all x, we must have A= 0 and B= 1/2.`
Were you trying to use the "variation of parameters" method? That works but, here, is extremely tedious and error-prone. In this case the "right hand side" is sin(x) which is one of the kinds of functions we expect to get as a solution to a "linear equation with constant coefficients". When the "right hand side" is an exponential, a polynomial, sine or cosine, or combinations of those, try "undetermined coefficients" rather than "variation of parameters".