The graphs $y=2 \cos 3 x+1$ and $y=-\cos 2 x$ intersect at many points. Two of these points, $P$ and $Q$, have $x$-coordinates between $\frac{17 \pi}{4}$ and $\frac{21 \pi}{4}$. The line through $P$ and $Q$ intersects the $x$-axis at $B$ and the $y$-axis at $A$. If $O$ is the origin, what is the area of $\triangle B O A ?$
We find the points of intersection of $y=2 \cos 3 x+1$ and $y=-\cos 2 x$ by equating values of $y$ and obtaining the following equivalent equations:

$$
\begin{aligned}
2 \cos 3 x+1 & =-\cos 2 x \\
2 \cos (2 x+x)+\cos 2 x+1 & =0 \\
2(\cos 2 x \cos x-\sin 2 x \sin x)+\cos 2 x+1 & =0 \\
2\left(\left(2 \cos ^{2} x-1\right) \cos x-(2 \sin x \cos x) \sin x\right)+\left(2 \cos ^{2} x-1\right)+1 & =0 \\
2\left(2 \cos ^{3} x-\cos x-2 \sin ^{2} x \cos x\right)+2 \cos ^{2} x & =0 \\
4 \cos ^{3} x-2 \cos x-4\left(1-\cos ^{2} x\right) \cos x+2 \cos ^{2} x & =0 \\
4 \cos ^{3} x-2 \cos x-4 \cos x+4 \cos ^{3} x+2 \cos ^{2} x & =0 \\
8 \cos ^{3} x+2 \cos ^{2} x-6 \cos x & =0 \\
4 \cos ^{3} x+\cos ^{2} x-3 \cos x & =0 \\
\cos x\left(4 \cos ^{2} x+\cos x-3\right) & =0 \\
\cos x(\cos x+1)(4 \cos x-3) & =0
\end{aligned}
$$

Therefore, $\cos x=0$ or $\cos x=-1$ or $\cos x=\frac{3}{4}$.

Two of these points of intersection, $P$ and $Q$, have $x$-coordinates between $\frac{17 \pi}{4}=4 \pi+\frac{\pi}{4}$ and $\frac{21 \pi}{4}=5 \pi+\frac{\pi}{4}$.

Since $\cos 4 \pi=1$ and $\cos \left(4 \pi+\frac{\pi}{4}\right)=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}=\frac{2 \sqrt{2}}{4}<\frac{3}{4}$, then there is an angle $\theta$ between $4 \pi$ and $\frac{17 \pi}{4}$ with $\cos \theta=\frac{3}{4}$ and so there is not such an angle between $\frac{17 \pi}{4}$ and $\frac{21 \pi}{4}$.

Therefore, we look for angles $x$ between $\frac{17 \pi}{4}$ and $\frac{21 \pi}{4}$ with $\cos x=0$ and $\cos x=-1$.

Note that $\cos 5 \pi=-1$ and that $\cos \frac{18 \pi}{4}=\cos \frac{9 \pi}{2}=\cos \frac{\pi}{2}=0$.

Thus, the $x$-coordinates of $P$ and $Q$ are $5 \pi$ and $\frac{9 \pi}{2}$.

Suppose that $P$ has $x$-coordinate $5 \pi$.

Since $P$ lies on $y=-\cos 2 x$, then its $y$-coordinate is $y=-\cos 10 \pi=-1$.

Suppose that $Q$ has $x$-coordinate $\frac{9 \pi}{2}$.

Since $Q$ lies on $y=-\cos 2 x$, then its $y$-coordinate is $y=-\cos 9 \pi=1$.

Therefore, the coordinates of $P$ are $(5 \pi,-1)$ and the coordinates of $Q$ are $\left(\frac{9 \pi}{2}, 1\right)$.

The slope of the line through $P$ and $Q$ is $\frac{1-(-1)}{\frac{9 \pi}{2}-5 \pi}=\frac{2}{-\frac{\pi}{2}}=-\frac{4}{\pi}$.

The line passes through the point with coordinates $(5 \pi,-1)$.

Thus, its equation is $y-(-1)=-\frac{4}{\pi}(x-5 \pi)$ or $y+1=-\frac{4}{\pi} x+20$ or $y=-\frac{4}{\pi} x+19$.

This line intersects the $y$-axis at $A(0,19)$. Thus, $A O=19$.

This line intersects the $x$-axis at point $B$ with $y$-coordinate 0 and hence the $x$-coordinate of $B$ is $x=19 \cdot \frac{\pi}{4}=\frac{19 \pi}{4}$. Thus, $B O=\frac{19 \pi}{4}$.

Since $\triangle A O B$ is right-angled at the origin, then its area equals $\frac{1}{2}(A O)(B O)=\frac{1}{2}(19)\left(\frac{19 \pi}{4}\right)=\frac{361 \pi}{8}$.