<p>Values of \(x\) satisfying \(\sin^{-1}(x^2-5x+7)=2\tan^{-1}1\):</p>
Step-by-Step Solution
<div class="solution"><p><strong>Step 1:</strong> $2\tan^{-1}(1)=2\cdot\pi/4=\pi/2$.</p><p><strong>Step 2:</strong> $\sin^{-1}(x^2-5x+7)=\pi/2\implies x^2-5x+7=1$.</p><p><strong>Step 3:</strong> $x^2-5x+6=0\implies(x-2)(x-3)=0\implies x=2,3$.</p><p><strong>Answer: (A),(B) \to x=2 and x=3</strong></p><div class="trap-box"><strong>Trap:</strong> Forgetting to verify the argument stays in [-1,1] -- here it's forced to 1, so valid.<div class="key-concept"><strong>Key Concept:</strong> Map the constant RHS to its unit-circle value first
Correct Answer: A,B