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

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

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 \(0 \leq x \leq 2\pi\). 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 \(0 \leq x \leq 2\pi\) are \(x = 0, \pi/3, 5\pi/3, 2\pi\). In other words, \(x = \boxed{0, \pi/3, 5\pi/3, 2\pi}\)