The set of values of k for which `x^2 - kx + sin^-1 (sin 4) > 0` for all real x is

Answer

To solve the inequality \( x^2 - kx + \sin^{-1}(\sin 4) > 0 \) for all real \( x \), we will follow these steps:

### Step 1: Simplify \(\sin^{-1}(\sin 4)\)

First, we need to simplify \(\sin^{-1}(\sin 4)\). The sine function is periodic, and we can express \(4\) in terms of a reference angle. Since \(4\) is greater than \(\pi\) (approximately \(3.14\)), we can use the identity:

\[ \sin 4 = \sin(\pi - 4) \]

Thus, we have:

\[ \sin^{-1}(\sin 4) = \pi - 4 \]

### Step 2: Rewrite the inequality

Now, we can rewrite the original inequality:

\[ x^2 - kx + (\pi - 4) > 0 \]

### Step 3: Analyze the quadratic inequality

For the quadratic \(x^2 - kx + (\pi - 4)\) to be greater than \(0\) for all real \(x\), its discriminant must be less than \(0\). The discriminant \(D\) of a quadratic \(ax^2 + bx + c\) is given by:

\[ D = b^2 - 4ac \]

In our case, \(a = 1\), \(b = -k\), and \(c = \pi - 4\). Therefore, the discriminant is:

\[ D = (-k)^2 - 4 \cdot 1 \cdot (\pi - 4) = k^2 - 4(\pi - 4) \]

### Step 4: Set the discriminant less than zero

To ensure the quadratic is positive for all \(x\), we need:

\[ k^2 - 4(\pi - 4) < 0 \]

This simplifies to:

\[ k^2 < 4(\pi - 4) \]

### Step 5: Solve for \(k\)

Taking the square root of both sides gives us:

\[ |k| < 2\sqrt{\pi - 4} \]

This means:

\[ -2\sqrt{\pi - 4} < k < 2\sqrt{\pi - 4} \]

### Step 6: Determine the values of \(\pi - 4\)

Since \(\pi \approx 3.14\), we find:

\[ \pi - 4 \approx 3.14 - 4 = -0.86 \]

Since \(\pi - 4\) is negative, \(2\sqrt{\pi - 4}\) is not a real number. Therefore, there are no real values of \(k\) that satisfy this inequality.

### Conclusion

The set of values of \(k\) for which the quadratic \(x^2 - kx + \sin^{-1}(\sin 4) > 0\) for all real \(x\) is empty.