QPaperGen

We are given:
\[
f(x) = \log \cos x
\]

Differentiate with respect to \( x \):
\[
f'(x) = \frac{d}{dx} \left( \log \cos x \right) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x
\]

Now analyze the sign of \( f'(x) = -\tan x \) in the given intervals:

\[\] In \( \left(0, \dfrac{\pi}{2} \right) \),
\( \tan x > 0 \Rightarrow -\tan x < 0 \Rightarrow f'(x) < 0 \)
\( \Rightarrow f \) is strictly decreasing on \( \left(0, \dfrac{\pi}{2} \right) \).

\[\] In \( \left( \dfrac{\pi}{2}, \pi \right) \),
\( \tan x < 0 \Rightarrow -\tan x > 0 \Rightarrow f'(x) > 0 \)
\( \Rightarrow f \) is strictly increasing on \( \left( \dfrac{\pi}{2}, \pi \right) \).
\[\]
\(\textbf{Conclusion:}\)
\[
f(x) = \log \cos x \text{ is strictly decreasing on } \left(0, \dfrac{\pi}{2} \right) \text{ and strictly increasing on } \left( \dfrac{\pi}{2}, \pi \right).
\]