Don't hit and try!

Let $\color{#3D99F6}{y=2^{2x-1}}$

From equation,

$\begin{aligned}2^{2x-1}+2^{1-2x}=&\ 2\\2^{2x-1}+2^{-(2x-1)}=&\ 2\\\color{#3D99F6}{2^{2x-1}}+(\color{#3D99F6}{2^{2x-1}})^{-1}=&\ 2\\\color{#3D99F6}y+\dfrac1{\color{#3D99F6}y}=&\ 2\\y^2+1=&\ 2y\\y^2-2y+1=&\ 0\\(y-1)^2=&\ 0\\\color{#3D99F6}{y=}&\ \color{#3D99F6}1\end{aligned}$

Thus,

$\begin{aligned}\color{#3D99F6}{2^{2x-1}=}&\ \color{#3D99F6}1\\2^{\color{#D61F06}{2x-1}}=&\ 2^{\color{#D61F06}0}\\\color{#D61F06}{2x-1}=&\ \color{#D61F06}0\\x=&\ \dfrac12=\boxed{0.5}\end{aligned}$

$2^ {2x-1}+2^ {1-2x}=2$ $\Rightarrow{\dfrac {2 ^{2x}}{2}+\dfrac {2}{2^{2x}}=2}$ Squaring both sides,we have, $\dfrac {2^{4x}}{4}+\dfrac {4}{2^{4x}}+2=4$ $\Rightarrow {\dfrac {2^{4x}}{4}+\dfrac {4}{2^{4x}}-2=0}$ $\Rightarrow {\left (\dfrac {2^{2x}}{2}-\dfrac {2}{2^{2x}}\right)^{2}=0}$ $\Rightarrow {\dfrac{2^{2x}}{2}-\dfrac {2}{2^{2x}}=0}$ $\Rightarrow {\dfrac {2^{2x}}{2}=\dfrac {2}{2^{2x}}}$ Cross-multiplying, we get, $2^ {4x}=2^2$ $\Rightarrow {4x=2}$ \color\green{\boxed {x=0.5}}