Simplifying the Function

The given function is \(f(x) = 2\cos^2x - 1\). Using the double angle identity for cosine, \(\cos(2x) = 2\cos^2x - 1\), we can simplify the function to:

\(f(x) = \cos(2x)\)

Finding the Derivative

To determine where the function is decreasing, we first find its derivative, \(f'(x)\).

Using the chain rule, the derivative of \(f(x) = \cos(2x)\) is:

\(f'(x) = -\sin(2x) \cdot \frac{d}{dx}(2x)\)

\(f'(x) = -\sin(2x) \cdot 2\)

\(f'(x) = -2\sin(2x)\)

Identifying Decreasing Intervals

A function is decreasing where its derivative is negative (\(f'(x) < 0\)). Therefore, we need to solve the inequality:

\(-2\sin(2x) < 0\)

Divide both sides by -2 and reverse the inequality sign:

\(\sin(2x) > 0\)

The sine function, \(\sin(\theta)\), is positive when \(\theta\) lies in the first and second quadrants. This occurs when:

\(2k\pi < 2x < \pi + 2k\pi\), where \(k\) is any integer.

Divide the inequality by 2 to find the intervals for \(x\):

\(k\pi < x < \frac{\pi}{2} + k\pi\)

Calculating the Longest Interval Length

The intervals where the function \(f(x)\) is decreasing are of the form \((k\pi, k\pi + \frac{\pi}{2})\).

The length of any such interval is calculated as the difference between the upper and lower bounds:

Length \(= (\frac{\pi}{2} + k\pi) - (k\pi)\)

Length \(= \frac{\pi}{2}\)

Since the length is \(\frac{\pi}{2}\) for all integer values of \(k\), the length of the longest interval in which the function is decreasing is \(\frac{\pi}{2}\).