Example 2. Solve the inequality

$$
x^{2}-2|x|<3
$$
Solution. The given inequality is equivalent to the combination of two systems:

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } - 2 x - 3 < 0 , } \\
{ x \geqslant 0 , }
\end{array} \left\{\begin{array}{l}
x^{2}+2 x-3<0 \\
x<0
\end{array}\right.\right.
$$

Since $x^{2}-2 x-3=(x+1)(x-3)$, the set of all solutions to the inequality $x^{2}-2 x-3<0$ is the interval $-1<x<3$; therefore, the solution to the first system of the combination is the interval $0 \leqslant x<3$.

From the equality $x^{2}+2 x-3=(x+3)(x-1)$, it follows that the inequality $x^{2}+2 x-3<0$ is satisfied only for $-3<x<1$. Hence, the solution to the second system of the combination is the interval $-3<x<0$.

By combining the solution sets of the two systems, we obtain the solution set of the inequality (2) - the interval $-3<x<3$.

The inequality (2) can also be solved by substituting the variable $t=|x|$: first, find the solution to the system

$$
\left\{\begin{array}{l}
t^{2}-2 t<3 \\
t \geqslant 0
\end{array}\right.
$$

i.e., the interval $0 \leqslant t<3$, and then solve the inequality $0 \leqslant|x|<3$.

As a result, we obtain the solutions to the inequality (2): $-3<x<3$.