Given, $\cos 2 x+2 \sin x=2 \alpha-7$

$\Rightarrow 1-2 \sin ^{2} x+\alpha \sin x=2 \alpha-7$

$\Rightarrow 2 \sin ^{2} x-\alpha \sin x+2 \alpha-8=0$

$\Rightarrow \sin x=\frac{\alpha \pm \sqrt{\alpha^{2}-8(2 \alpha-8)}}{4}$

$\Rightarrow \sin x=\frac{\alpha \pm(\alpha-8)}{4}$

$\Rightarrow \sin x=\frac{\alpha+\alpha-8}{4}, \frac{\alpha-\alpha+8}{4}$

$\sin x=2$ (Not possible)

For solution $-1 \leq \frac{2 \alpha-8}{4} \leq 1$

$-4 \leq 2 \alpha-8 \leq 4$

$\Rightarrow 4 \leq 2 \alpha \leq 12$

$\Rightarrow \alpha \in[2,6]$