h(y) = 2y + \frac{1}{2y}

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• Let $ y = \sin^2 x $, where $ y > 0 $. Then:
• g(x) = 2y + 2 + \frac{1}{2y}
• We minimize $ 2y + \frac{1}{2y} + 2 $. Let:
• Using the AM-GM inequality:
• 2y + \frac{1}{2y} \geq 2\sqrt{2y \cdot \frac{1}{2y}} = 2
• Equality holds when $ 2y = \frac{1}{2y} \Rightarrow 4y^2 = 1 \Rightarrow y = \frac{1}{2} $.

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