jmc

Subtracting $1$ from both sides, we get $$-\frac{1}{2}\le \frac{1}{x}\le \frac{1}{2}.$$Note that we cannot take the reciprocal of all the quantities to solve for $x,$ because the quantities do not have the same signs. Instead, we consider the two inequalities $-\frac{1}{2}\le \frac{1}{x}$ and $\frac{1}{x}\le \frac{1}{2}$ separately. Break into cases on the sign of $x.$ If $x>0,$ then $-\frac{1}{2}\le \frac{1}{x}$ is always true, and the inequality $\frac{1}{x}\le \frac{1}{2}$ implies $x\ge 2.$ If $x<0,$ then $\frac{1}{x}\le \frac{1}{2}$ is always true, and the inequality $-\frac{1}{2}\le \frac{1}{x}$ implies $x\le -2.$ Hence, the solution set is $$x\in \boxed{(-\infty ,-2]\cup [2,\infty )}.$$