a) Ta có : [imath]cos x \in [-1;1] \longrightarrow 0 \le \sqrt{cosx} \le 1 \longrightarrow 1 \le 2\sqrt{cosx} + 1 \le 3[/imath]
Vậy:min = 1, xảy ra khi [imath]cosx=0 \longrightarrow x = \frac{\pi}{2} + k\pi (k \in Z)[/imath]
max=3, xảy ra khi [imath]cosx=1 \longrightarrow x = 2k\pi (k \in Z)[/imath]
b)Ta có : [imath]sin x \in [-1;1] \longrightarrow -1 \le sinx \le 1 \longrightarrow -2 \le -2sinx \le 2 \longrightarrow 1 \le 3-2sinx \le 5[/imath]
Vậy: Min = 1 xảy ra khi [imath]sinx=1 hay x = \dfrac{\pi}{2} + 2k\pi (k \in Z)[/imath]
Max = 5 xảy ra khi [imath]sinx=-1 hay x = -\dfrac{\pi}{2} + 2k\pi (k \in Z)[/imath]