The Problems of Ukrainian Authors

Answer: $x>0$.

Easy to understand, that range of a function in our expression is set $x>0$. We will show that this is the solution. Really when $x>0$ $2008^x>1$, ${\log }_{2008}(x+1)>0$, because under such conditions, the logarithm base satisfies the conditions $0<\frac{x}{x+2}<1$, ${\sin }^{2007}x+{\cos }^{2008}x\le {\sin }^{2}x+{\cos }^{2}x=1$. Thus the left part is greater than 1, and right - lower. Therefore the solution of inequality is every point $x>0$.