`a)`

Vì `-1 \le \cos x \le 1`, nên:

`->\min A = 4 + 3.(-1) = 1 `

`->\max A = 4 + 3.1 = 7`

`b)`

Vì `0 \le \sin^2 x \le 1`, nên:

`->\min B = \frac{3 - 2.1}{5} = \frac{1}{5} `

`->\max B = \frac{3 - 2.0}{5} = \frac{3}{5}`

`c)`

Ta có : `C = 4(1 - \sin^2 x) - 2\sin x - 6 = -4\sin^2 x - 2\sin x - 2`

Đặt `t = \sin x \in [-1, 1]->f(t) = -4t^2 - 2t - 2 `

Đạo hàm: `f'(t)-8t-2`

`f'(t)=0<=>t=-1/4`

`->f(-\frac{1}{4}) = -4(-\frac{1}{4})^2 - 2(-\frac{1}{4}) - 2 = -4 . \frac{1}{16} + \frac{1}{2} - 2`

`= -\frac{1}{4} + \frac{1}{2} - 2 = -\frac{7}{4}`

`->f(-1) = -4.1 - 2.(-1) - 2 = -4 + 2 - 2 = -4`

`->f(1) = -4.1 - 2.1 - 2 = -4 - 2 - 2 = -8`

 Vậy : `GT``N``N = -8 `

`GTLN= -\frac{7}{4} `