Official solution
1 - 2sin2x + \[\alpha \] sin x = 2\[\alpha \] - 7
\[ \Rightarrow \] 2sin2x - \[\alpha \] 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
\[ - 1 \le {{\alpha - 4} \over 2} \le 1 \Rightarrow - 2 \le \alpha - 4 \le 2 \Rightarrow 2 \le \alpha \le 6\]
\[ \therefore \] \[\alpha \in \left[ {2,6} \right]\]