1 - 2sin 2 x + $\alpha$ sin x = 2$\alpha$ - 7 $\Rightarrow$ 2sin 2 x - $\alpha$ sin x + (2$\alpha$ - 8) = 0 $\sin x = {{\alpha \pm \sqrt {{\alpha ^2} - 8(2\alpha - 8)} } \over 4}$ $\Rightarrow {{\alpha \pm \sqrt {{\alpha ^2} - 16\alpha + 64} } \over 4} = {{\alpha \pm (\alpha - 8)} \over 4}$ ${{\alpha \pm (\alpha - 8)} \over 4} \Rightarrow {{2\alpha - 8} \over 4},2$ $\Rightarrow {{\alpha - 4} \over 2},2$ (rejected) To exists solutions $- 1 \le {{\alpha - 4} \over 2} \le 1 \Rightarrow - 2 \le \alpha - 4 \le 2 \Rightarrow 2 \le \alpha \le 6$ $\therefore$ $\alpha \in \left[ {2,6} \right]$