number of solutions..

\hspace{-16}$Here $\mathbf{\mid [x]-2x\mid =4}$\\\\ Means $\mathbf{[x]-2x=4}$ and $\mathbf{[x]-2x=-4}$\\\\ \bullet\;\; $\mathbf{When\; [x]-2x=4}$\\\\ Put $\mathbf{x=I+f}\;,$ where $\mathbf{0\leq f<1}$\\\\ $\mathbf{I-2(I+f)=4\Leftrightarrow f=-\left(\frac{I+4}{2}\right)}$\\\\ Now $\mathbf{0\leq f<1\Leftrightarrow 0\leq -\left(\frac{I+4}{2}\right)<1}$\\\\ So $\mathbf{-6<I\leq -4}$\\\\ Now $\mathbf{I\in\mathbb{Z}}\;,$ So $\mathbf{I=-5\;,-4}$\\\\ So If $\mathbf{I=-5}\;,$ Then $\mathbf{I=-\left(\frac{-5+4}{2}\right)=\frac{1}{2}}$\\\\ So $\mathbf{x=I+f=-5+\frac{1}{2}=-\frac{9}{2}}$\\\\ So $\mathbf{\boxed{\bold{x=-\frac{9}{2}}}}$\\\\ Similarly If $\mathbf{I=-4}\;,$ Then $\mathbf{I=-\left(\frac{-4+4}{2}\right)=0}$\\\\ So $\mathbf{x=I+f=-4+0=-4}$\\\\ So $\mathbf{\boxed{\bold{x=-4}}}$\\\\

\hspace{-16}\bullet\;\; \mathbf{When\; [x]-2x=-4}$\\\\ Put $\mathbf{x=I+f}\;,$ where $\mathbf{0\leq f<1}$\\\\ $\mathbf{I-2(I+f)=-4\Leftrightarrow f=\left(\frac{4-I}{2}\right)}$\\\\ Now $\mathbf{0\leq f<1\Leftrightarrow 0\leq \left(\frac{4-I}{2}\right)<1\Leftrightarrow 2<I\leq 4}$\\\\ Now $\mathbf{I\in\mathbb{Z}}\;,$ So $\mathbf{I=3\;,4}$\\\\ So If $\mathbf{I=3}\;,$ Then $\mathbf{I=\left(\frac{4-3}{2}\right)=\frac{1}{2}}$\\\\ So $\mathbf{x=I+f=3+\frac{1}{2}=\frac{7}{2}}$\\\\ So $\mathbf{\boxed{\bold{x=\frac{7}{2}}}}$\\\\ Similarly If $\mathbf{I=4}\;,$ Then $\mathbf{I=\left(\frac{4-4}{2}\right)=0}$\\\\ So $\mathbf{x=I+f=4+0=4}$\\\\ So $\mathbf{\boxed{\bold{x=4}}}$\\\\ So $\mathbf{x\in \left\{-\frac{9}{2}\;,-4\;,\frac{7}{2}\;,4\right\}}$

|2x-[x]| = |2[x]+2\{x \}-[x]| = |[x]+2\{x \}|=4 \Rightarrow 2\{x \} \in \mathbb{Z}

\Rightarrow \{x \} = 0 \ \text{or} \ \frac{1}{2}

So, if {x} =0, we get x=±4. If {x} = 1/2, we get

\left|x+ \frac{1}{2}\right| = 4 \Rightarrow x = \frac{7}{2} \ \text{or} \ x=-\frac{9}{2}