Solve $2 \sin ^{2}(w)+3 \sin (w)+1=0$ for all solutions $0 \leq w< 2 \pi$.

Give your answers as exact values in a list separated by commas.

Step 1 ：Understand the problem: We are asked to solve the equation \(2 \sin ^{2}(w)+3 \sin (w)+1=0\) for all solutions \(0 \leq w<2 \pi\).

Step 2 ：Identify the type of equation: This is a quadratic equation in terms of \(\sin(w)\).

Step 3 ：Let \(x = \sin(w)\). Then the equation becomes \(2x^2 + 3x + 1 = 0\).

Step 4 ：Solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 3\), and \(c = 1\). Substituting these values into the formula, we get: \(x = \frac{-3 \pm \sqrt{3^2 - 4*2*1}}{2*2} = \frac{-3 \pm \sqrt{9 - 8}}{4} = \frac{-3 \pm 1}{4}\). So, the solutions for \(x\) are \(x = -1\) and \(x = -0.5\).

Step 5 ：Since \(x = \sin(w)\), we need to find the values of \(w\) such that \(\sin(w) = -1\) and \(\sin(w) = -0.5\). For \(\sin(w) = -1\), the solution is \(w = \pi + \pi/2 = 3\pi/2\). For \(\sin(w) = -0.5\), the solutions are \(w = \pi + \pi/6 = 7\pi/6\) and \(w = \pi + 5\pi/6 = 11\pi/6\).

Step 6 ：Check the solutions: All these solutions are in the interval \(0 \leq w < 2 \pi\), so they are valid.

Step 7 ：\(\boxed{\text{The solutions to the equation } 2 \sin ^{2}(w)+3 \sin (w)+1=0 \text{ for } 0 \leq w<2 \pi \text{ are } w = 3\pi/2, 7\pi/6, 11\pi/6}\)