2. If $\frac{1-\cos \theta}{4+\sin ^{2} \theta}=\frac{1}{2}$, then $\left(4+\cos ^{3} \theta\right) \cdot\left(3+\sin ^{3} \theta\right)=$
2. 9 Detailed Explanation: From the condition, we get $2-2 \cos \theta=4+\sin ^{2} \theta, \Rightarrow(\cos \theta-1)^{2}=4, \Rightarrow \cos \theta=-1$, then $\left(4+\cos ^{3} \theta\right) \cdot\left(3+\sin ^{3} \theta\right)=9$.