3. Given $\log _{10} \sin x+\log _{10} \cos x=-1$. Then $2 \log _{10}(\sin x+\cos x)=$ $\qquad$
3. $\lg 6-\lg 5$.

Notice that, $\sin x, \cos x$ are both positive, and
$$
\begin{array}{l}
\sin x \cdot \cos x=\frac{1}{10} . \\
\text { Then } 2 \log _{10}(\sin x+\cos x) \\
=\log _{10}(\sin x+\cos x)^{2} \\
=\log _{10}(1+2 \sin x \cdot \cos x) \\
=\log _{10} \frac{6}{5}=\lg 6-\lg 5 .
\end{array}
$$