The set of all real values of x for which the funciton `f(x) = sqrt(sin x + cos x )+sqrt(7x -x^(2) - 6)` takes real values is

Explanation

To find the set of all real values of \( x \) for which the function

\[ f(x) = \sqrt{\sin x + \cos x} + \sqrt{7x - x^2 - 6} \]

takes real values, we need to ensure that both terms under the square roots are non-negative.

### Step 1: Analyze the first term \(\sqrt{\sin x + \cos x}\)

The expression inside the first square root must be non-negative:

\[ \sin x + \cos x \geq 0 \]

We can rewrite \(\sin x + \cos x\) using the angle addition formula:

\[ \sin x + \cos x = \sqrt{2} \left( \sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}} \right) = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]

Thus, we need:

\[ \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \geq 0 \]

This implies:

\[ \sin\left(x + \frac{\pi}{4}\right) \geq 0 \]

The sine function is non-negative in the intervals:

\[ n\pi \leq x + \frac{\pi}{4} \leq (n+1)\pi \quad \text{for } n \in \mathbb{Z} \]

This simplifies to:

\[ n\pi - \frac{\pi}{4} \leq x \leq (n+1)\pi - \frac{\pi}{4} \]

### Step 2: Analyze the second term \(\sqrt{7x - x^2 - 6}\)

The expression inside the second square root must also be non-negative:

\[ 7x - x^2 - 6 \geq 0 \]

Rearranging gives:

\[ -x^2 + 7x - 6 \geq 0 \]

Factoring the quadratic:

\[ -(x^2 - 7x + 6) \geq 0 \implies (x - 1)(x - 6) \leq 0 \]

The roots of this quadratic are \( x = 1 \) and \( x = 6 \). The solution to the inequality \( (x - 1)(x - 6) \leq 0 \) is:

\[ 1 \leq x \leq 6 \]

### Step 3: Combine the results

Now we need to find the intersection of the intervals obtained from both conditions:

1. From \(\sin x + \cos x \geq 0\), we have intervals of the form \(n\pi - \frac{\pi}{4} \leq x \leq (n+1)\pi - \frac{\pi}{4}\).
2. From \(7x - x^2 - 6 \geq 0\), we have \(1 \leq x \leq 6\).

To find the specific intervals for \(n\), we can evaluate:

- For \(n = 0\):
\[ 0 - \frac{\pi}{4} \leq x \leq \pi - \frac{\pi}{4} \implies -\frac{\pi}{4} \leq x \leq \frac{3\pi}{4} \]

- For \(n = 1\):
\[ \pi - \frac{\pi}{4} \leq x \leq 2\pi - \frac{\pi}{4} \implies \frac{3\pi}{4} \leq x \leq \frac{7\pi}{4} \]

Now we need to check which of these intervals overlap with \(1 \leq x \leq 6\):

- The interval \(-\frac{\pi}{4} \leq x \leq \frac{3\pi}{4}\) overlaps with \(1 \leq x \leq \frac{3\pi}{4}\).
- The interval \(\frac{3\pi}{4} \leq x \leq \frac{7\pi}{4}\) overlaps with \(1 \leq x \leq 6\) since \(\frac{7\pi}{4} > 6\).

### Final Answer

Thus, the set of all real values of \( x \) for which the function \( f(x) \) takes real values is:

\[ [1, \frac{3\pi}{4}] \cup [\frac{3\pi}{4}, 6] \]