Both curves complete one loop over the interval [tex]0\le\theta\le2\pi[/tex]. See the attached plot. First find when the curves intersect in this interval:
[tex]2-2\sin\theta=3\implies\sin\theta=-\dfrac12\implies\theta=\dfrac{7\pi}6,\theta=\dfrac{11\pi}6[/tex]
From the plot we can see that [tex]r=2-2\sin\theta[/tex] is "larger" than [tex]r=3[/tex], so the area is given by
[tex]\displaystyle\int_{7\pi/6}^{11\pi/6}((2-2\sin\theta)-3)\,\mathrm d\theta=-\int_{7\pi/6}^{11\pi/6}(1+2\sin\theta)\,\mathrm d\theta[/tex]
[tex]\displaystyle=\int_{11\pi/6}^{7\pi/6}(1+2\sin\theta)\,\mathrm d\theta[/tex]
[tex]=(\theta-2\cos\theta)\bigg|_{\theta=11\pi/6}^{\theta=7\pi/6}[/tex]
[tex]=\left(\dfrac{7\pi}6-2\cos\dfrac{7\pi}6\right)-\left(\dfrac{11\pi}6-2\cos\dfrac{11\pi}6\right)[/tex]
[tex]=2\sqrt3-\dfrac{2\pi}3[/tex]
