$$ \begin{aligned} & \log _{10} \sin x+\log _{10} \cos x=-1, x \in(0, \pi / 2) \\\\ & \log _{10}(\sin x \cos x)=-1 \\\\ & \Rightarrow \sin x \cos x=10^{-1}= {1 \over {10}} \\\\ & \log _{10}(\sin x+\cos x)={1 \over {2}}\left(\log _{10} n-1\right), n>0 \\\\ & 2 \log _{10}(\sin x+\cos x)=\left(\log _{10} n-\log _{10} 10\right) \\\\ & \Rightarrow \log _{10}(\sin x+\cos x)^2=\log _{10}({n \over {10}}) \\\\ & \Rightarrow (\sin x+\cos x)^2={n \over {10}} \\\\ & \Rightarrow \sin ^2 x+\cos ^2 x+2 \sin x \cos x=\frac{n}{10} \\\\ & \Rightarrow 1+2({1 \over {10}})={n \over {10}} \Rightarrow {12 \over {10}}={n \over {10}} \\\\ & \therefore n=12 \end{aligned} $$