Use ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$, ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$, $2\sin \theta \cos \theta =\sin 2\theta$.

$f(\theta )=\frac{1+\cos 2\theta }{2}-3\sin 2\theta +3\cdot \frac{1-\cos 2\theta }{2}+2=4-3\sin 2\theta -\cos 2\theta$.

$3\sin 2\theta +\cos 2\theta$ has amplitude $\sqrt{9+1}=\sqrt{10}$, so it lies in $[-\sqrt{10},\sqrt{10}]$.

$\therefore f(\theta )\in [4-\sqrt{10},4+\sqrt{10}]$. Least value $=4-\sqrt{10}$.