Solution: Let $ y = \frac{2\sin x + 1}{\sin x - 3} $. Solve for $ \sin x $: $ y(\sin x - 3) = 2\sin x + 1 \Rightarrow y\sin x - 3y = 2\sin x + 1 \Rightarrow \sin x(y - 2) = 3y + 1 $. Thus, $ \sin x = \frac{3y + 1}{y - 2} $. Since $ |\sin x| \leq 1 $, solve $ \left| \frac{3y + 1}{y - 2} \right| \leq 1 $. This leads to $ -1 \leq \frac{3y + 1}{y - 2} \leq 1 $. Sol

Mon, 04 May 2026 23:43:49 +0000

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