Помогите решить уравнение |x-|x-|x-1|||=1/2

$\big |x-|x-|x-1||\big |=\frac{1}{2}\\\\a)\; \; x \geq 1\; ,\; \; |x-1|=x-1\; \; \to \; \big |x-|x-(x-1)|\big |=\frac{1}{2}\; ,\\\\\big |x-|1|\big |=\frac{1}{2}\; ,\; \to \; |x-1|=\frac{1}{2}\; ,\; \to \; x-1=\pm \frac{1}{2}\\\\x-1=\frac{1}{2}\; ,\; \to \; x_1=\frac{3}{2} \geq 1\\\\x-1=-\frac{1}{2},\; \to \; \; x_2=\frac{1}{2} \ \textless \ 1\; (ne\; podxodit)\\\\b)\; \; x\ \textless \ 1\; ,|x-1|=1-x\; \; \to \; \; \big |x-|x-(1-x)|\big |=\frac{1}{2}\\\\\big |x-|2x-1|\big |=\frac{1}{2}$

$Esli\; x\ \textless \ 1,\; to\; \; 2x\ \textless \ 2\; \; i\; \; 2x-1\ \textless \ 1\; \; \to \\\\b_1)\; \; 0 \leq 2x-1\ \textless \ 1\\\\\big |x-|2x-1|\big |=\big |x-2x+1\big |=\big |1-x\big |=\frac{1}{2}\\\\1-x=\frac{1}{2}\; \; \; ili\; \; \; 1-x=-\frac{1}{2}\\\\x=\frac{1}{2}\ \textless \ 1\; \; \; ili\; \; \; x=\frac{3}{2}\ \textgreater \ 1\; (ne\; podxodit)\\\\b_2)\; \; 2x-1\ \textless \ 0\; \; \to \; \; |2x-1|=1-2x\\\\\big |x-(1-2x)\big |=\frac{1}{2}\\\\\big |3x-1\big |=\frac{1}{2}$

$3x-1=\frac{1}{2}\; \; \; ili\; \; \; 3x-1=-\frac{1}{2}\\\\x=\frac{1}{2}\ \textless \ 1\; \; \; \; ili\; \; \; \; x=\frac{1}{6}\ \textless \ 1$

Ответ: $x=\frac{1}{6}\; ,\; \; x=\frac{1}{2}\; ,\; \; x=\frac{3}{2}$