$$\begin{array}{l}8{\sin ^2}x + 3\cos 2x\\ = 4\left( {1 - \cos 2x} \right) + 3\cos 2x\\ = 3\cos 2x - 4\cos 2x + 4\\ = 4 - \cos 2x\\ - 1 \le \cos 2x \le 1 \Rightarrow  - 1 \le  - \cos 2x \le 1\\ \Leftrightarrow 3 \le 4 - \cos 2x \le 5\\ \Rightarrow T = \left[ {3;5} \right]\end{array}$$

1) 
y = 8sin^{2} x + 3cos 2x
  = 8.\frac{1 - cos 2x}{2} + 3cos 2x
  = 4(1 - cos 2x) + 3cos 2x
  = 4 - 4cos 2x + 3cos 2x
  = 4 - cos 2x = 0
\text{Ta có:} - 1 le cos 2x le 11
          <=> 1 \ge - cos 2x \ge - 1
          <=> 5 \ge 4 - cos 2x \ge 3
          <=> 5 \ge y \ge 3
=> T = [3;5]
Vậy ___________________________________________________
$$\color{pink}{\text{$$\textit{$$\circ$$ hungnguyen4269}$$}}$$