5. Given the vector $\boldsymbol{a}=(\cos \theta, \sin \theta)$, vector $\boldsymbol{b}=(\sqrt{3},-1)$, then the maximum value of $|2 \boldsymbol{a}-\boldsymbol{b}|$ is
5. 4 Hint: $2 \boldsymbol{a}-\boldsymbol{b}=(2 \cos \theta-\sqrt{3} \cdot 1+2 \sin \theta) \quad$ so $\quad|2 \boldsymbol{a}-\boldsymbol{b}|=\sqrt{8-8 \cos \left(\theta+\frac{\pi}{6}\right)}$ Therefore, when $\cos \left(\theta+\frac{\pi}{6}\right)=-1$, $|2 \boldsymbol{a}-\boldsymbol{b}|_{\text {max }}=4$