Plot the curves with -π/2 ≤ θ ≤ π/2.
The two curves intersect when
7 cos(θ) = 3 + cos(θ) ⇒ 6 cos(θ) = 3
⇒ cos(θ) = 1/2
⇒ θ = arccos(1/2) + 2nπ or θ = -arccos(1/2) + 2nπ
⇒ θ = ± π/3 + 2nπ
where n is an integer; with θ in the prescribed interval, intersections occur for θ = ± π/3.
Then the area of the region is
[tex]\displaystyle \int_{\theta=-\pi/3}^{\pi/3} \int_{3+\cos(\theta)}^{7\cos(\theta)} r \, dr \, \dtheta \\\\ = \frac12 \int_{-\pi/3}^{\pi/3} \left((7\cos(\theta))^2 - (3+\cos(\theta))^2\right) \, d\theta \\\\ = \frac12 \int_{-\pi/3}^{\pi/3} \left(48\cos^2(\theta) - 9 - 6\cos(\theta)\right) \, d\theta[/tex]
Since the integrand is even, the integral over a symmetric interval [-k, k] is equal to twice the integral over half the interval [0, k] :
[tex]\displaystyle \int_0^{\pi/3} \left(48\cos^2(\theta) - 6\cos(\theta) - 9\right) \, d\theta[/tex]
Recall the half-angle identity for cosine,
[tex]\cos^2(\theta) = \dfrac{1 + \cos(2\theta)}2[/tex]
to simplify the integral to
[tex]\displaystyle \int_0^{\pi/3} \left(1 - 2\cos(\theta) + 6\cos(2\theta)\right) \, d\theta[/tex]
and we evaluate this to be
[tex]\displaystyle \left(\theta - 2\sin(\theta) + 3 \sin(2\theta)\right) \bigg|_{\theta=0}^{\pi/3} = \frac\pi3 - 2\sin\left(\frac\pi3\right) + 3\sin\left(\frac{2\pi}3\right) \\\\ = \frac\pi3 + \sin\left(\frac\pi3\right) = \boxed{\frac\pi3 + \frac{\sqrt3}2}[/tex]
