Question:

<div class="solution"><p><strong>Step 1:</strong> $5\in(3\pi/2,2\pi)$ so $\sin^{-1}(\sin 5)=5-2\pi$.</p><p><strong>Step 2:</strong> Inequality: $x^2-4x+(2\pi-5)<0$.</p><p><strong>Step 3:</strong> Roots: $x=2\pm\sqrt{9-2\pi}$.</p><p><strong>Answer: (A) $x\in(2-\sqrt{9-2\pi},\,2+\sqrt{9-2\pi})$</strong></p><div class="trap-box"><strong>Trap:</strong> Taking $\sin^{-1}(\sin 5)=5$ -- wrong because 5 is outside the principal range $[-\pi/2,\pi/2]$.<div class="key-concept"><strong>Key Concept:</strong> Always reduce angle to principal branch before solving ITF inequalities