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 $0 < \frac{x}{x+2} < 1$, $\sin^{2007}x + \cos^{2008}x \le \sin^2x + \cos^2x = 1$. Thus the left part is greater than 1, and right - lower. Therefore the solution of inequality is every point $x > 0$.