3. Given $\boldsymbol{a}=\left(\lambda+2, \lambda^{2}-\cos ^{2} \alpha\right), \boldsymbol{b}=\left(m, \frac{m}{2}+\sin \alpha\right)$, where $\lambda, m, \alpha$ are real numbers, if $\boldsymbol{a}=2 \boldsymbol{b}$, then the range of $\frac{\lambda}{m}$ is $\qquad$ .
According to the problem, $\left\{\begin{array}{l}\lambda+2=2 m, \\ \lambda^{2}-\cos ^{2} \alpha=2\left(\frac{m}{2}+\sin \alpha\right)\end{array} \Rightarrow(2 m-2)^{2}-m=\cos ^{2} \alpha+2 \sin \alpha \Rightarrow 4 m^{2}-9 m+4\right.$ $=-(\sin \alpha-1)^{2}+2 \in[-2,2] \Rightarrow m \in\left[\frac{1}{4}, 2\right]$. Therefore, $\frac{\lambda}{m}=2-\frac{2}{m} \in[-6,1]$.