Trigonometry problem of jee level

Let us rewrite the above trig equation as:

$1-2sin^{2}x + \alpha sin x = 2\alpha - 7 \Rightarrow 0 = 2sin^{2}x - \alpha sin x + (2\alpha - 8)$ ;

or $sin x = \frac{\alpha \pm \sqrt{\alpha^{2} - 4(2)(2\alpha-8)}}{4} = \frac{\alpha \pm \sqrt{\alpha^{2} -16\alpha + 64}}{4} = \frac{\alpha \pm \sqrt{(\alpha - 8)^2}}{4} = \frac{\alpha \pm (\alpha - 8)}{4} = 2, \frac{\alpha}{2} - 2.$

Since $-1 \le sin x\le 1,$ 2 cannot be a solution. Hence we require: $-1 \le \frac{\alpha}{2} - 2 \le 1 \Rightarrow \boxed{2 \le \alpha \le 6}.$ This gives us 5 possible integral solutions for $\alpha.$