$f(x)=-\sin x+2\cos x+3$
$=-\sqrt5.\Big(\dfrac{1}{\sqrt5}\sin x-\dfrac{2}{\sqrt5}\cos x\Big)+3$
Đặt $\cos\alpha=\dfrac{1}{\sqrt5};\sin\alpha=\dfrac{2}{\sqrt5}$
$\to f(x)=-\sqrt5\sin(x-\alpha)+3$
Ta có: $-1\le \sin(x-\alpha)\le 1$
$\to -\sqrt5\le -\sqrt5\sin(x-\alpha)\le \sqrt5$
$\to 3-\sqrt5\le f(x)\le 3+\sqrt5$
Vậy $\min f(x)=3-\sqrt5$