solve `sin^(-1) (sin 5) gt x^(2) - 4x`

Explanation

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

### Step 1: Simplifying \( \sin^{-1}(\sin 5) \)

The function \( \sin^{-1}(x) \) returns values in the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Since \( 5 \) is outside this range, we need to find an equivalent angle within this range.

Using the property of the sine function, we know:
\[ \sin y = \sin x \implies y = n\pi + (-1)^n x \]
for some integer \( n \).

Since \( 5 \) is in the first quadrant, we can express it as:
\[ \sin^{-1}(\sin 5) = 5 - 2n\pi \]
where \( n \) is chosen such that \( 5 - 2n\pi \) lies within \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

### Step 2: Finding the appropriate \( n \)

To find \( n \), we need to check values of \( n \):
- For \( n = 0 \):
\[ 5 - 2(0)\pi = 5 \quad (\text{not in range}) \]
- For \( n = 1 \):
\[ 5 - 2(1)\pi = 5 - 2\pi \approx 5 - 6.28 = -1.28 \quad (\text{in range}) \]
- For \( n = 2 \):
\[ 5 - 2(2)\pi = 5 - 4\pi \quad (\text{not in range}) \]

Thus, the valid expression is:
\[ \sin^{-1}(\sin 5) = 5 - 2\pi \]

### Step 3: Setting up the inequality

Now we substitute back into the inequality:
\[ 5 - 2\pi > x^2 - 4x \]

Rearranging gives:
\[ x^2 - 4x + (2\pi - 5) < 0 \]

### Step 4: Solving the quadratic inequality

Let \( f(x) = x^2 - 4x + (2\pi - 5) \). We need to find the roots of the equation:
\[ x^2 - 4x + (2\pi - 5) = 0 \]

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -4, c = 2\pi - 5 \):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (2\pi - 5)}}{2 \cdot 1} \]
\[ x = \frac{4 \pm \sqrt{16 - 8\pi + 20}}{2} \]
\[ x = \frac{4 \pm \sqrt{36 - 8\pi}}{2} \]
\[ x = 2 \pm \sqrt{9 - 2\pi} \]

### Step 5: Finding the interval

The roots are:
\[ x_1 = 2 - \sqrt{9 - 2\pi}, \quad x_2 = 2 + \sqrt{9 - 2\pi} \]

The quadratic opens upwards (since the coefficient of \( x^2 \) is positive), thus \( f(x) < 0 \) between the roots:
\[ x \in (2 - \sqrt{9 - 2\pi}, 2 + \sqrt{9 - 2\pi}) \]

### Final Answer

The solution to the inequality \( \sin^{-1}(\sin 5) > x^2 - 4x \) is:
\[ x \in (2 - \sqrt{9 - 2\pi}, 2 + \sqrt{9 - 2\pi}) \]