Question:

Step 1: Evaluate $\sin^{-1}(\sin 5)$ The principal value range for $\sin^{-1}(y)$ is $[-\pi/2, \pi/2]$. We need to find an angle $\theta \in [-\pi/2, \pi/2]$ such that $\sin \theta = \sin 5$. We know that $\pi \approx 3.14159$ and $3\pi/2 \approx 4.71239$. Since $5$ radians is in the interval $(\pi, 3\pi/2)$, $\sin 5$ is negative. The reference angle for $5$ in the interval $(\pi, 3\pi/2)$ is $5-\pi$. However, $5-\pi \approx 5 - 3.14159 = 1.85841$, which is not in $[-\pi/2, \pi/2]$. The angle $\theta$ such that $\sin \theta = \sin 5$ and $\theta \in [-\pi/2, \pi/2]$ is given by $\theta = 2\pi - 5$. Let's verify: $2\pi - 5 \approx 2(3.14159) - 5 = 6.28318 - 5 = 1.28318$. This value $1.28318$ is in the interval $[-\pi/2, \pi/2] \approx [-1.5708, 1.5708]$. Thus, $\sin^{-1}(\sin 5) = 2\pi - 5$. Step 2: Set up the inequality The given inequality is $\sin^{-1}(\sin 5) > x^2 - 4x$. Substituting the value from Step 1, we get: $$2\pi - 5 > x^2 - 4x$$ Step 3: Rearrange to standard form Rearranging the inequality, we obtain: $$x^2 - 4x < 2\pi - 5$$ $$x^2 - 4x - (2\pi - 5) < 0$$ Step 4: Complete the square To solve the quadratic inequality, we complete the square for the left side: $$x^2 - 4x + 4 - 4 - (2\pi - 5) < 0$$ $$(x - 2)^2 - 4 - 2\pi + 5 < 0$$ $$(x - 2)^2 + 1 - 2\pi < 0$$ $$(x - 2)^2 < 2\pi - 1$$ Step 5: Solve the inequality Taking the square root of both sides: $$|x - 2| < \sqrt{2\pi - 1}$$ This implies: $$-\sqrt{2\pi - 1} < x - 2 < \sqrt{2\pi - 1}$$ Adding 2 to all parts of the inequality: $$2 - \sqrt{2\pi - 1} < x < 2 + \sqrt{2\pi - 1}$$ The inequality holds for $x \in (2 - \sqrt{2\pi - 1}, 2 + \sqrt{2\pi - 1})$.