<s>[INST] Find the number of roots of the equation $3^{|x|}-|2-|x||=1$ 
Find the number of roots of the equation $3^{|x|}-|2-|x||=1$     

My working: 
Let $t$ be any positive real number.
$3^{t}-|2-t|=1$     
Case 1:
$t<2$
$3^{t}-2+t=1$
$3^{t}+t=3$     
Case 2:
$t>2$
$3^{t}+2-t=1$
$3^t+1=t$      
Now I don't know how to proceed further to solve these equations. I would require a hint for that.
 [/INST] $f(x)=3^{|x|}-|2-|x||$ is an even function ,so  
$f(x)=1$ has symmetric roots (like $x=\pm a$) 
so ,solve the equation for $x \geq 0 $ and then add negative of root(s)
$$3^{|x|}-|2-|x||=1,  x \geq 0 \\ 3^x-|2-x|=1 \to 
 \begin{cases}3^x-(2-x)=1 & x <2\\3^x+(2-x)=1 & x >2\end{cases}\\
\begin{cases}3^x=3-x & x <2\\3^x=-1-x & x >2\end{cases}\\
\begin{cases}3^x=3-x & x <2\\3^x=-1-x &  x >2 \end{cases}$$
first one has a solution , but second one has not root because $x>2 \to -1-x<-3\\3^x>0$
finally the original equation has 2 roots 
another observation is to take $3^{|x|}=1+|2-|x||$ there is $f(x)=g(x)$ plot them together ,so must be there two cross section between $f(x)$ and $g(x)$

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