Here is a solution. It uses the intersections library, Because of this I drew the sine curves with the builtin sin and cos paths; the intersections library does not work well with function plots.
The intersections library computes the intersection point of the two curves; you don't have to fiddle around trying to figure it out yourself.
The code is not elegant, the paths are repeated quite often, but it gets the job done.
\documentclass[border=5pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{intersections}
\begin{document}
\begin{tikzpicture}
% Achsen zeichnen
\draw[->,thick] (0,0) -- (10,0) node[below] {$x$};
\draw[->,thick] (0,0) -- (0,3.5) node[left] {$y$};
% Achsen beschriften
\foreach \x/\xlabel in {1/1,3.141/\pi,6.283/{2\pi}}
\draw (\x,-.2) -- ++(0,0.2) node[below=4pt] {$\scriptstyle\xlabel$};
\foreach \y in {0,1,2}
\draw (-.1,\y) -- ++(0.1,0) node[left=4pt] {$\scriptstyle\y$};
% solvent
\path[name path = solvent] (0,0) cos (0.5*pi,1) sin (pi,2) cos (1.5*pi,1) sin (2*pi,0);
%unsolvent
\path[shift={(1.2*pi,0)},name path= unsolvent] (0,0) cos (0.5*pi,1) sin (pi,2) cos (1.5*pi,1) sin (2*pi,0);
\begin{scope}
\clip[name intersections={of=solvent and unsolvent}] (intersection-1 |- {(0,0)}) rectangle (2*pi,2);
\fill[blue] (0,0) cos (0.5*pi,1) sin (pi,2) cos (1.5*pi,1) sin (2*pi,0);
\end{scope}
\begin{scope}
\clip[name intersections={of=solvent and unsolvent}] (0,0) rectangle (intersection-1 |- {(0,2)});
\fill[shift={(1.2*pi,0)},red] (0,0) cos (0.5*pi,1) sin (pi,2) cos (1.5*pi,1) sin (2*pi,0);
\end{scope}
% solvent
\draw[blue] (0,0) cos (0.5*pi,1) sin (pi,2) cos (1.5*pi,1) sin (2*pi,0);
%unsolvent
\draw[shift={(1.2*pi,0)},red] (0,0) cos (0.5*pi,1) sin (pi,2) cos (1.5*pi,1) sin (2*pi,0);
%cutoff
\draw [thick,green, name intersections={of=solvent and unsolvent}]
(intersection-1 |- {(0,0)}) -- ++(0,2) node[above] {cut off};
\end{tikzpicture}
\end{document}
The result is: