[tex]\it 2\{x\}-[x]=\dfrac{1}{4}\Big|_{\cdot4} \Rightarrow 8\{x\}-4[x]=1 \Rightarrow 8(x-[x])-4[x]=1 \Rightarrow \\ \\ \Rightarrow 8x-8[x]-4[x]=1 \Rightarrow 8x-12[x]=1 \Rightarrow 12[x]=8x-1 \Rightarrow 8x-1=k\in\mathbb{Z} \Rightarrow \\ \\ \Rightarrow x=\dfrac{k+1}{8}\ \ \ \ \ (1)[/tex]
Ecuația devine:
[tex]\it 12\Big[\dfrac{k+1}{8}\Big]=k \Rightarrow \Big[\dfrac{k+1}{8}\Big]=\dfrac{k}{12}\ \ \ \ \ (2)\\ \\ \\ (2) \Rightarrow k\in M_{12}\ \ \ \ \ (3)\\ \\ \\(2) \Rightarrow \dfrac{k}{12}\leq\dfrac{k+1}{8}<\dfrac{k}{12}+1\Big|_{\cdot48} \Rightarrow 4k\leq6k+6<4k+48|_{-4k} \Rightarrow \\ \\ \\ \Rightarrow 0\leq2k+6<48|_{-6} \Rightarrow -6\leq2k<42|_{:2} \Rightarrow -3\leq k<21\ \ \ \ \ (4)[/tex]
[tex]\it (3),\ (4) \Rightarrow k\in\{0,\ \ 12\}\ \ \ \ \ (5)\\ \\ \\(1),\ (5) \Rightarrow x\in\Big\{\dfrac{1}{8},\ \dfrac{13}{8}\Big\}[/tex]
