What is the length of the longest interval in which the function...

Given $f(x) = 2\cos^2 x - 1$

Using the double angle formula, we can write $f(x) = \cos 2x$

To find the intervals where the function is decreasing, we find its derivative and set it to be less than or equal to zero:

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

For $f(x)$ to be decreasing, $f'(x) \le 0$

$-2\sin 2x \le 0$

$\sin 2x \ge 0$

The sine function is positive in the first and second quadrants, so:

$2n\pi \le 2x \le 2n\pi + \pi$

Dividing by $2$, we get:

$n\pi \le x \le n\pi + \dfrac{\pi}{2}$

The length of the longest interval in which the function is decreasing is the difference between the upper and lower bounds of this interval:

Length $= \left(n\pi + \dfrac{\pi}{2}\right) - n\pi = \dfrac{\pi}{2}$

Answer: $\dfrac{\pi}{2}$