Solve the following equation in the set of real numbers:

$$
\sqrt{\lg ^{2} x+\lg x^{2}+1}+\lg x+1=0
$$
According to the definition of logarithm, $\lg x$ is defined if $x>0$.

The expression under the square root, $\lg ^{2} x+2 \lg x+1=(\lg x+1)^{2}$, can be written in this form, which is always non-negative, so the square root always has a meaning. Therefore, we look for the solution in the set of positive numbers.

By definition, $\sqrt{a^{2}}=|a|$. Using this, (1) can be written as:

$$
|\lg x+1|=-(\lg x+1)
$$

Since $|a|=+a$ if $a \geq 0$, and $-a$ if $a<0$, we need to distinguish two cases.

1. If $\lg x+1 \geq 0$, then the equation is

$$
\lg x+1=-\lg x-1
$$

from which

$$
\lg x=-1, \quad x=\frac{1}{10}
$$

By substitution, we can verify that this is indeed a root.

2. If $\lg x+1<0$, then our equation is

$$
-(\lg x+1)=-(\lg x+1)
$$

We arrive at an identity - provided that our initial inequality $(\lg x+1<0)$ holds. That is, $\lg x<-1$, from which $0<x<\frac{1}{10}$ follows.

Therefore, the solution is: $0<x \leq \frac{1}{10}$.