Answer: \(\frac{\sqrt3}{3}-\frac{\pi}{9}\).

The points of intersection satisfy

\(2+\cos3x=1.5.\)

Thus

\(\cos3x=-\frac12.\)

For the shaded region shown,

\(3x=\frac{2\pi}{3},\quad \frac{4\pi}{3}.\)

So

\(x=\frac{2\pi}{9},\quad \frac{4\pi}{9}.\)

In this interval, the line lies above the curve. Hence the area is

\(\int_{2\pi/9}^{4\pi/9}\left(1.5-(2+\cos3x)\right)\,dx.\)

So

\(\text{Area}=\int_{2\pi/9}^{4\pi/9}\left(-\frac12-\cos3x\right)\,dx.\)

Integrating,

\(\text{Area}=\left[-\frac12x-\frac13\sin3x\right]_{2\pi/9}^{4\pi/9}.\)

Substitute the limits:

\(\left(-\frac{2\pi}{9}-\frac13\sin\frac{4\pi}{3}\right)-\left(-\frac{\pi}{9}-\frac13\sin\frac{2\pi}{3}\right).\)

Since \(\sin\frac{4\pi}{3}=-\frac{\sqrt3}{2}\) and \(\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}\), this becomes

\(-\frac{\pi}{9}+\frac{\sqrt3}{6}+\frac{\sqrt3}{6}.\)

Therefore the exact area is

\(\boxed{\frac{\sqrt3}{3}-\frac{\pi}{9}}.\)