Step-by-step Solution:

To find the number of real solutions, we first simplify both sides of the equation and then analyze the resulting equation within the specified interval.

The given equation is \(\sqrt{1+\cos{2x}}=\sqrt{2}\cos^{-1}(\cos(x))\) for \(x \in [\frac{\pi}{2}, \pi]\).

1. Simplify the Left-Hand Side (LHS):
We use the double-angle identity for cosine, \(1+\cos(2x) = 2\cos^2(x)\).
\[LHS = \sqrt{1+\cos(2x)} = \sqrt{2\cos^2(x)} = \sqrt{2}\sqrt{\cos^2(x)}\]
It is crucial to remember that \(\sqrt{y^2} = |y|\). Therefore, \(\sqrt{\cos^2(x)} = |\cos(x)|\).
\[LHS = \sqrt{2}|\cos(x)|\]
2. Simplify the Right-Hand Side (RHS):
The property of the inverse cosine function states that \(\cos^{-1}(\cos(x)) = x\) if and only if \(x\) is within the principal value range of \([0, \pi]\).
The interval given in our problem, \([\frac{\pi}{2}, \pi]\), is entirely contained within this principal range. Thus, we can simplify the expression directly.
\[RHS = \sqrt{2}\cos^{-1}(\cos(x)) = \sqrt{2}x\]
3. Formulate the Simplified Equation:
By setting the simplified LHS equal to the simplified RHS, we get:
\[\sqrt{2}|\cos(x)| = \sqrt{2}x\]
Dividing both sides by \(\sqrt{2}\) yields:
\[|\cos(x)| = x\]
4. Analyze the Equation in the Given Interval \([\frac{\pi}{2}, \pi]\):
In the second quadrant, which corresponds to the interval \([\frac{\pi}{2}, \pi]\), the value of the cosine function is non-positive (i.e., \(\cos(x) \le 0\)).
Therefore, the absolute value \(|\cos(x)|\) simplifies to \(-\cos(x)\).
Our equation becomes:
\[-\cos(x) = x \quad \text{or} \quad \cos(x) = -x\]
5. Check for Solutions:
We need to determine if there is any value of \(x\) in \([\frac{\pi}{2}, \pi]\) for which \(\cos(x) = -x\). Let's examine the values of both functions in this interval.
- For \(y=\cos(x)\), the range of values is from \(\cos(\frac{\pi}{2})=0\) down to \(\cos(\pi)=-1\). So, \(-1 \le \cos(x) \le 0\).
- For \(y=-x\), the range of values is from \(-\frac{\pi}{2}\) down to \(-\pi\). Using \(\pi \approx 3.14\), this is approximately \(-1.57\) to \(-3.14\).

Comparing the ranges, the minimum value of \(\cos(x)\) is -1, while the maximum value of \(-x\) is approximately -1.57. Since \(-1 > -1.57\), the value of \(\cos(x)\) is always greater than the value of \(-x\) in the given interval. The two functions never intersect.

Result:
There are no real solutions to the equation in the interval \([\frac{\pi}{2}, \pi]\). The number of solutions is 0.