1. If the real number $x$ satisfies $\lg \sin x+\lg \cos x=-1$, then $\lg (\sin x+\cos x)=$
1. $\frac{1}{2}(\lg 12-1)$.

From $\lg \sin x+\lg \cos x=\lg (\sin x \cos x)=-1$, we get $\sin x \cos x=\frac{1}{10}$. Also,
$$
\begin{array}{l}
(\sin x+\cos x)^{2}=\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=\frac{12}{10} \\
\Rightarrow 2 \lg (\sin x+\cos x)=\lg \frac{12}{10}=\lg 12-1 \\
\Rightarrow \lg (\sin x+\cos x)=\frac{1}{2}(\lg 12-1) .
\end{array}
$$