`cos^(-1) ""(cos"" (7pi )/(6))`

Answer

To solve the problem \(\cos^{-1} (\cos (7\pi / 6))\), we need to simplify the expression and find the angle within the principal range of the inverse cosine function, which is \([0, \pi]\).

### Step-by-Step Solution:

1. **Understanding the Principal Range:**
The principal range of \(\cos^{-1} x\) is \([0, \pi]\). This means that the output of the \(\cos^{-1}\) function must lie within this interval.

2. **Express \(7\pi / 6\) in Terms of \(\pi\):**
Notice that \(7\pi / 6\) is greater than \(\pi\). We can express \(7\pi / 6\) as:
\[ 7\pi / 6 = \pi + \pi / 6 \]
This shows that \(7\pi / 6\) is in the third quadrant.

3. **Cosine in the Third Quadrant:**
In the third quadrant, the cosine function is negative. Therefore:
\[ \cos(7\pi / 6) = \cos(\pi + \pi / 6) = -\cos(\pi / 6) \]

4. **Simplify Using the Cosine Function:**
We know that:
\[ \cos(\pi / 6) = \frac{\sqrt{3}}{2} \]
Therefore:
\[ \cos(7\pi / 6) = -\frac{\sqrt{3}}{2} \]

5. **Apply the Inverse Cosine Function:**
Now, we need to find \(\cos^{-1}(-\frac{\sqrt{3}}{2})\). The angle whose cosine is \(-\frac{\sqrt{3}}{2}\) within the principal range \([0, \pi]\) is:
\[ \cos^{-1}(-\frac{\sqrt{3}}{2}) = \pi - \cos^{-1}(\frac{\sqrt{3}}{2}) \]

6. **Evaluate \(\cos^{-1}(\frac{\sqrt{3}}{2})\):**
We know that:
\[ \cos^{-1}(\frac{\sqrt{3}}{2}) = \pi / 6 \]
Therefore:
\[ \cos^{-1}(-\frac{\sqrt{3}}{2}) = \pi - \pi / 6 = 5\pi / 6 \]

### Final Answer:
\[ \cos^{-1} (\cos (7\pi / 6)) = 5\pi / 6 \]