First, by the definition of logarithm function, $x$ must be positive. Then, $\log_{\frac{1}{2}}x\ne 0\implies x\ne 1$. Hence, the basic requirement for $x$ is $$x\in (0, 1) \cup (1, +\infty)$$ There are two possible cases: - case 1: $\frac{1}{\log_{\frac{1}{2}}x}+2 >\frac{3}{2}$ - case 2: $\frac{1}{\log_{\frac{1}{2}}x}+2 < -\frac{3}{2}$ $\underline{case\ 1:}$ $$\frac{1}{\log_{\frac{1}{2}}x}+2 >\frac{3}{2}\implies\frac{1}{\log_{\frac{1}{2}}x} > -\frac{1}{2}$$ If $\log_{\frac{1}{2}}x > 0$ or, equivalently, $x < 1$, this inequality always hold. If $\log_{\frac{1}{2}}x < 0$ or, equivalently, $x > 1$, we have $$\log_{\frac{1}{2}}x < -2 \implies x > 4$$ $\underline{case\ 2:}$ $$\frac{1}{log_{\frac{1}{2}}x}+2 < - \frac{3}{2} \implies \frac{1}{\log_{\frac{1}{2}}x} < -\frac{7}{2}\implies\log_{\frac{1}{2}}x > -\frac{2}{7}\implies x < 2^{\frac{2}{7}}$$ Combining these results leads to the final answer:$$\boxed{(0,1)\cup(1, 2^{\frac{2}{7}})\cup(4,+\infty)}$$