For the function $F(x)=\left|\cos ^{2} x+2 \sin x \cos x-\sin ^{2} x+A x+B\right|$, when $x \in \left[0, \frac{3 \pi}{2}\right]$, the maximum value $M$ is related to the parameters $A$ and $B$. What values of $A$ and $B$ minimize $M$? Prove your conclusion.
Distribute in $M$ is the maximum value under the premise, construct inequality relations through special values, and combine with the intuitive analysis of function graphs.
Solution 1 (1) If $A=B=0, F(x)=\left|\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)\right|$, then when $x=\frac{\pi}{8}$, $\frac{5 \pi}{8}$, $\frac{9 \pi}{8}$, the maximum value $M$ of $F(x)$ is $\sqrt{2}$.
(2) If $A=0, B \neq 0, F(x)=\left|\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)+B\right|$, at this time
$$
M=\max \{|\sqrt{2}+B|,|\sqrt{2}-B|\}>\sqrt{2} .
$$
(3) If $A \neq 0, B=0, F(x)=\left|\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)+A x\right|$. If $A>0$, $F\left(\frac{\pi}{8}\right)=\left|\sqrt{2}+A \cdot \frac{\pi}{8}\right|>\sqrt{2}$, at this time $M>\sqrt{2}$; if $A<0$, $F\left(\frac{9 \pi}{8}\right)=\left|\sqrt{2}+A \cdot \frac{9 \pi}{8}\right|>\sqrt{2}$, at this time there is also $M>\sqrt{2}$.
(4) If $A \neq 0, B \neq 0$, as shown in Figure 9-1, the line $y=A x+B$ must have a part in the first or fourth quadrant, intersecting with at least one of the rays $l_{1}$, $l_{2}$, $l_{3}$. At the intersection point, the function values of $y=A x+B$ and $y=\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)$ have the same sign, and the absolute value of their sum must be less than $\sqrt{2}$, so there is also $M>\sqrt{2}$. Consider. When $A=0$ it is the same as above, when $A \neq 0$ the discussion is as follows:
(1) If $\frac{5 \pi}{8} A+B<0$, then $F\left(\frac{5 \pi}{8}\right)=\left|\sqrt{2}+\frac{5 \pi}{8} A+B\right|>\sqrt{2}$;
(2) If $\frac{5 \pi}{8} A+B \geqslant 0, \frac{9 \pi}{8} A+B=\left(\frac{5 \pi}{8} A+B\right)+\frac{4 \pi}{8} A$ and $\frac{\pi}{8} A+B=$ $\left(\frac{5 \pi}{8} A+B\right)-\frac{4 \pi}{8} A$ at least one is greater than 0, i.e., $F\left(\frac{9 \pi}{8}\right)>\sqrt{2}$ or $F\left(\frac{\pi}{8}\right)>\sqrt{2}$ at least one holds, so there is always $M>\sqrt{2}$. Therefore, only when $A=B=0$, $M=\sqrt{2}$, in other cases there is always $M>\sqrt{2}$.