The equation $a\sin x+\cos 2x=2a-7$  possesses a solution if

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

$\therefore 2{\sin }^{2}x-a\sin x+2a-8=0$

$\sin x=\frac{a\pm \sqrt{a^2-\mathrm{4.2.}\left(2a-8\right)}}{4}$

$\sin x=\frac{a\pm \left(a-8\right)}{4}$

$-1\le \frac{a+a-8}{4}\le 1$    (or)  $-1\le \frac{a-a-8}{4}\le 1$

$2\le a\le 6$

$-1\le \frac{a+a-8}{4}\le 1$  it is not possible