Solve $2 \cos ^{2}(w)-3 \cos (w)+1=0$ for all solutions. \[ w=\square+2 k \pi \text {; Select an answer (0) } \] Give your answers as exact values in a list separated by commas Question Help: $\square$ Video $\square$ Message instructor Submit Question

Step 1 ：First, we need to solve the quadratic equation \(2 \cos ^{2}(w)-3 \cos (w)+1=0\). We can do this by using the quadratic formula, which gives us the roots of the equation.

Step 2 ：The roots of the quadratic equation are the possible values for \(\cos(w)\). We find that the roots are 1 and 0.5.

Step 3 ：Next, we use the arccosine function to find the corresponding values of \(w\). The arccosine of 1 is 0, and the arccosine of 0.5 is \(\frac{\pi}{3}\) (or approximately 1.047).

Step 4 ：However, these are not the only solutions. Since the cosine function has a period of \(2\pi\), we can add \(2k\pi\) to these solutions to get the general solutions, where \(k\) is an integer.

Step 5 ：The general solutions are \(w = \boxed{0 + 2k\pi}\) and \(w = \boxed{\frac{\pi}{3} + 2k\pi}\), where \(k\) is an integer.