Solve for all values of $x$ that satisfy the following equation: $2{\cos }^2(x)-3\cos (x)+1=0$ where $0\pi \le x\le 2\pi$

Step 1 ：Given the equation $2{\cos }^2(x)-3\cos (x)+1=0$ where $0\le x\le 2\pi$

Step 2 ：This is a quadratic equation in terms of $\cos (x)$. We can solve it by using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a=2$, $b=-3$, and $c=1$

Step 3 ：Calculate the discriminant $D=b^2-4ac=1$

Step 4 ：Find the roots of the equation using the quadratic formula. The roots are $1.0$ and $0.5$

Step 5 ：Find the corresponding $x$ values within the given range $0\le x\le 2\pi$. The $x$ values are $0.0$, $6.283185307179586$, $1.0471975511965979$, $5.235987755982988$

Step 6 ：Simplify the $x$ values to $0$, $2\pi$, $\pi /3$, $5\pi /3$

Step 7 ：Final Answer: The solutions to the equation $2{\cos }^2(x)-3\cos (x)+1=0$ within the range $0\le x\le 2\pi$ are $x=0,\pi /3,5\pi /3,2\pi$. In other words, $x=\boxed{{\displaystyle 0,\pi /3,5\pi /3,2\pi }}$