Solution: Let $ y = \frac2\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 = \frac3y + 1y - 2 $. Since $ |\sin x| \leq 1 $, solve $ \left| \frac3y + 1y - 2 \right| \leq 1 $. This leads to $ -1 \leq \frac3y + 1y - 2 \leq 1 $. Sol - Veritas Home Health