The minimum value of $e^{(2x^2-2x+1){\sin }^{2}x}$ is:

Let $f(x)=e^{(2x^2-2x+1){\sin }^{2}x}$.The exponent is $g(x)=(2x^2-2x+1){\sin }^{2}x$.Note that $2x^2-2x+1=2(x^2-x+1/2)=2[(x-1/2{)}^2+1/4]>0$ for all real $x$.Also, ${\sin }^{2}x\ge 0$ for all real $x$.Thus, $g(x)\ge 0$ for all $x$.The minimum value of $g(x)$ occurs when $\sin x=0$ (e.g., at $x=0$ or $x=\pi$).When $\sin x=0$, $g(x)=(2x^2-2x+1)\cdot 0=0$.The minimum value of the original function is $e^{g_{min}}=e^0=1$.