Solution: Rewrite $ f(x) = \frac{\sin^2 x + 4}{\sin x} = \sin x + \frac{4}{\sin x} $. Let $ y = \sin x \in (0, 1] $. The function becomes $ f(y) = y + \frac{4}{y} $. The derivative $ f'(y) = 1 - \frac{4}{y^2} $ has critical point at $ y = 2 $, but $ y \leq 1 $. Analyze endpoints: as $ y \to 0^+ $, $ f(y) \to \infty $; at $ y = 1 $, $ f(1) = 1 + 4 = 5 $. The minimum is $ 5 $.

Sat, 02 May 2026 12:11:17 +0000

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