3. Let $x$ be a real number, the maximum value of the function $f(x)=4 \cos ^{3} x-3 \cos ^{2} x-6 \cos x+5$ is


3. $\frac{27}{4}$.
$$
\begin{aligned}
f(x) & =4 \cos ^{3} x-3 \cos ^{2} x-6 \cos x+5=(1-\cos x)^{2}(4 \cos x+5) \\
& =\frac{1}{4}(2-2 \cos x)(2-2 \cos x)(4 \cos x+5) \\
& \leqslant \frac{1}{4}\left(\frac{(2-2 \cos x)+(2-2 \cos x)+(4 \cos x+5)}{3}\right)^{3} \\
& =\frac{27}{4} .
\end{aligned}
$$

The equality holds if and only if $2-2 \cos x=4 \cos x+5$, i.e., $\cos x=\frac{1}{2}$. Therefore, the maximum value of $f(x)$ is $\frac{27}{4}$.