5.2.3 Write down the minimum value of the function defined by \( y=\frac{\cos ^{4} x+\sin ^{2} x \cdot \cos ^{2} x}{1+\sin x} \)

To find the minimum value of the function \[ y = \frac{\cos^4 x + \sin^2 x \cos^2 x}{1 + \sin x}, \] we can start by simplifying the expression in the numerator. We know that \[ \cos^4 x = (\cos^2 x)^2, \] and we can represent \(\sin^2 x\) as \(1 - \cos^2 x\). This substitution gives us: \[ \sin^2 x \cos^2 x = (1 - \cos^2 x) \cos^2 x = \cos^2 x - \cos^4 x. \] Thus, the numerator becomes: \[ \cos^4 x + \sin^2 x \cos^2 x = \cos^4 x + \cos^2 x - \cos^4 x = \cos^2 x. \] Now substituting this back into the function, we have: \[ y = \frac{\cos^2 x}{1 + \sin x}. \] Next, we need to analyze this function. The range of \(\cos^2 x\) is from \(0\) to \(1\). The term \(1 + \sin x\) ranges from \(1\) (when \(\sin x = -1\)) to \(2\) (when \(\sin x = 1\)). Thus, the expression \(\frac{\cos^2 x}{1 + \sin x}\) will vary based on these bounds. To find the minimum value, we should consider critical points and the endpoints. Noticing that \(\sin x\) at its minimum (-1) gives: \[ y = \frac{\cos^2 x}{0}, \] which hints towards undefined behavior at that point. However, values closer to that will still be bound by \(0\). To analyze more carefully, we can explore specific values of \(x\): 1. When \(x = 0\): \[ y = \frac{\cos^2(0)}{1 + \sin(0)} = \frac{1}{1} = 1. \] 2. When \(x = \frac{\pi}{2}\): \[ y = \frac{\cos^2(\frac{\pi}{2})}{1 + \sin(\frac{\pi}{2})} = \frac{0}{2} = 0. \] 3. When \(x = \frac{\pi}{4}\): \[ y = \frac{\cos^2(\frac{\pi}{4})}{1 + \sin(\frac{\pi}{4})} = \frac{\left(\frac{1}{\sqrt{2}}\right)^2}{1 + \frac{1}{\sqrt{2}}} = \frac{\frac{1}{2}}{1 + \frac{1}{\sqrt{2}}}. \] By examining values and behavior, it appears the minimum is approached at \(x = \frac{\pi}{2}\), yielding: \[ \text{Minimum value of } y = 0. \] Thus, the minimum value of the function is \[ \boxed{0}. \]