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📅 May 15, 2026 👤 scraface

May 15, 2026

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e -2\) (which is always true since \(\sin x \in [-1,1]\)).
Solution: Let \(y = rac{2\sin x + 3}{\sin x + 2}\). Solve for \(\sin...
y(\sin x + 2) = 2\sin x + 3 \Rightarrow y\sin x + 2y = 2\sin x + 3
We require \(\left|rac{3 - 2y}{y - 2}
ight| \leq 1\)
\left| rac{3 - 2y}{y - 2}
ight| \leq 1 \Rightarrow (3 - 2y)^2 \leq (y - 2)^2
Expand both sides:
- 12y + 4y^2 \leq y^2 - 4y + 4
4y^2 - 12y + 9 \leq y^2 - 4y + 4

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