6.88 The function $F(x)=\left|\cos ^{2} x+2 \sin x \cos x-\sin ^{2} x+A x+B\right|$ has a maximum value $M$ on $0 \leqslant x \leqslant \frac{3}{2} \pi$ that depends on the parameters $A$ and $B$. For what values of $A$ and $B$ is $M$ minimized? Prove your conclusion.
[Solution] (1) $F(x)=\left|\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)+A x+B\right|$ When $A=B=0$, $F(x)$ becomes $f(x)=\sqrt{2}\left|\sin \left(2 x+\frac{\pi}{4}\right)\right|$. In the interval $\left[0, \frac{3}{2} \pi\right]$, there are three points $x_{1}=\frac{\pi}{8}, x_{2}=\frac{5 \pi}{8}, x_{3}=\frac{9 \pi}{8}$. At these points, $f(x)$ achieves the maximum value $M_{f}=\sqrt{2}$, which is the smallest value of $M$ required.
(2) Below, we prove that for any $A, B$ not both zero, we have
$$\max _{0 \leqslant r \leqslant \frac{3 \pi}{2}} F(x)>\max _{0 \leqslant x \leqslant \frac{3 \pi}{2}} f(x)=M_{f}=\sqrt{2}$$
(i) When $A=0, B \neq 0$, it is clear that
$$\max _{0 \leqslant x \leqslant \frac{3 \pi}{2}} F(x)=\max _{0 \leqslant x \leqslant \frac{3 \pi}{2}}\left|\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)+B\right|,$$

so (1) holds.
(ii) When $A>0, B \geqslant 0$,

Since $F\left(\frac{\pi}{8}\right)=\sqrt{2}+\frac{\pi}{8} A+B>\sqrt{2}$, (1) holds.
(iii) When $A>0, B<0$, we have two cases:
I. If $|B| < \frac{9 \pi}{8} A$, then $|B| < \frac{5 \pi}{8} A$, and $\frac{5 \pi}{8} A + B > -\sqrt{2}$, so
$$F\left(\frac{5 \pi}{8}\right)=\left|-\sqrt{2}+\frac{5 \pi}{8} A+B\right|>\sqrt{2},$$

so (1) holds.
II. If $|B| \geqslant \frac{9 \pi}{8} A$, then $|B|>\frac{5 \pi}{8} A$, and $\frac{5 \pi}{8} A + B < -\sqrt{2}$, so
$$F\left(\frac{5 \pi}{8}\right)=\left|-\sqrt{2}+\frac{5 \pi}{8} A+B\right|>\sqrt{2},$$

so (1) holds.
(iv) When $A<0, B \geqslant 0$,

Since $F\left(\frac{9 \pi}{8}\right)=\left|\sqrt{2}+\frac{9 \pi}{8} A+B\right|>\sqrt{2}$, (1) holds.
(v) When $A<0, B<0$, we have two cases:
I. If $B<-\frac{9 \pi}{8} A$, then $B<-\frac{5 \pi}{8} A$, and $\frac{5 \pi}{8} A + B < -\sqrt{2}$, so
$$F\left(\frac{5 \pi}{8}\right)=\left|-\sqrt{2}+\frac{5 \pi}{8} A+B\right|>\sqrt{2},$$

so (1) holds.
II. If $B \geqslant -\frac{5 \pi}{8} A$, then $B > -\frac{\pi}{8} A$, and $\frac{\pi}{8} A + B > 0$, so
$$F\left(\frac{\pi}{8}\right)=\left|\sqrt{2}+\frac{\pi}{8} A+B\right|>\sqrt{2}$$

so (1) holds.
In summary, (1) holds in all five cases.