`"sin"^(-1)("sin" 5)>x^2-4x` holds if

Options

A. `x=2-sqrt(9-2pi)`
B. `x=2+sqrt(9-2pi)`
C. `xgt2+sqrt(9-2pi)`
D. `x epsilon(2-sqrt(9-2pi),2+sqrt(9-2pi))`

Answer

To solve the inequality \( \sin^{-1}(\sin 5) > x^2 - 4x \), we can follow these steps:

### Step 1: Evaluate \( \sin^{-1}(\sin 5) \)
Since \( 5 \) is in the range \( \left( \frac{3\pi}{2}, 2\pi \right) \), we can find \( \sin 5 \) as follows:
\[ \sin 5 = \sin(5 - 2\pi) = \sin(5 - 6.2832) = \sin(-1.2832) \quad (\text{since } 2\pi \approx 6.2832) \]
Thus,
\[ \sin^{-1}(\sin 5) = 5 - 2\pi \]

### Step 2: Rewrite the inequality
Now we can rewrite the inequality:
\[ 5 - 2\pi > x^2 - 4x \]
This can be rearranged to:
\[ x^2 - 4x + (2\pi - 5) < 0 \]

### Step 3: Identify coefficients
Here, we identify the coefficients:
- \( a = 1 \)
- \( b = -4 \)
- \( c = 2\pi - 5 \)

### Step 4: Use the quadratic formula
The roots of the quadratic equation \( x^2 - 4x + (2\pi - 5) = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values:
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (2\pi - 5)}}{2 \cdot 1} \]
\[ x = \frac{4 \pm \sqrt{16 - 8\pi + 20}}{2} \]
\[ x = \frac{4 \pm \sqrt{36 - 8\pi}}{2} \]
\[ x = 2 \pm \sqrt{9 - 2\pi} \]

### Step 5: Determine the interval
The quadratic \( x^2 - 4x + (2\pi - 5) < 0 \) will be negative between its roots. Therefore, the solution to the inequality is:
\[ x \in \left( 2 - \sqrt{9 - 2\pi}, 2 + \sqrt{9 - 2\pi} \right) \]

### Final Answer
Thus, the inequality \( \sin^{-1}(\sin 5) > x^2 - 4x \) holds for:
\[ x \in \left( 2 - \sqrt{9 - 2\pi}, 2 + \sqrt{9 - 2\pi} \right) \]