How would you solve this?

A graph often helps in these sort of problems:

How would you end up solving, |2-|1-|x||| = 1,
i realize the basic idea but what do you do with the negative after the 2, do you distribute it?
edit: through the powers of trial and error i found
that you dont need to distribute but rather just divide the -1 and the answers are, +-4, +-2, and 0.

\(\begin{array}{|rcll|} \hline |~2-|~1-|x|~|~| &=& 1 \qquad & | \qquad \text{square both sides} \\ (~2-|~1-|x|~|~)^2 &=& 1^2 \\ 4-4\cdot |~1-|x|~| +(1-|x|)^2 &=& 1 \\ 4-4\cdot |~1-|x|~| + \not{1}-2|x| + x^2 &=& \not{1} \\ 4-4\cdot |~1-|x|~| -2|x| + x^2 &=& 0 \\ x^2-2|x|+4 &=& 4\cdot |~1-|x|~| \qquad & | \qquad \text{square both sides} \\ (x^2-2|x|+4)^2 &=& 16\cdot (~1-|x|~)^2 \\ (x^2-2|x|+4)\cdot (x^2-2|x|+4) &=& 16\cdot (~1 -2|x| + x^2 ~) \\ x^4-4|x|\cdot x^2+12x^2-16x+ \not \!\! 16 &=& \not \!\! 16 -32\cdot|x| + 16x^2 \\ x^4-4|x|\cdot x^2+12x^2-16x &=& -32\cdot|x| + 16x^2 \\ x^4-4|x|\cdot x^2+16|x|-4x^2 &=& 0 \\ 4|x|\cdot (4-x^2) &=& 4x^2-x^4 \\ 4|x|\cdot (4-x^2) &=& x^2\cdot(4-x^2) \qquad & | \qquad \text{square both sides} \\ 16x^2\cdot (4-x^2)^2 &=& x^4\cdot(4-x^2)^2 \\ 16x^2\cdot (4-x^2)^2 - x^4\cdot(4-x^2)^2 &=& 0 \\ (4-x^2)^2\cdot(16x^2 - x^4 ) &=& 0 \\ (4-x^2)^2\cdot x^2\cdot (16 - x^2 ) &=& 0 \\ (4-x^2)\cdot (4-x^2) \cdot x \cdot x\cdot (16 - x^2 ) &=& 0 \\ \hline \end{array} \)

1.

\(\begin{array}{|rcll|} \hline 4-x^2 &=& 0 \\ x^2 &=& 4 \\ x &=& \pm\sqrt{4} \\\\ \mathbf{x_{1}} &\mathbf{=} & \mathbf{2} \\ \mathbf{x_{2}} &\mathbf{=} & \mathbf{-2} \\ \hline \end{array} \)

2.

\(\begin{array}{|rcll|} \hline x &=& 0 \\ \Rightarrow \\ \mathbf{x_{3}} &\mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline 16-x^2 &=& 0 \\ x^2 &=& 16 \\ x &=& \pm\sqrt{16} \\\\ \mathbf{x_{4}} &\mathbf{=} & \mathbf{4} \\ \mathbf{x_{5}} &\mathbf{=} & \mathbf{-4} \\ \hline \end{array}\)