The set of values of a for which the equation $\cos 2x+a\sin x=3a-17$ possesses a solution is $\left[c,d\right]$ then d-c =
  

    $1-2{\sin }^{2}x+a\sin x=3a-17$

$\Rightarrow 2{\sin }^{2}x-a\sin x+3a-18=0$

$\Rightarrow \sin x=\frac{a\pm \sqrt{a^2-8\left(3a-18\right)}}{4}$

                  $$\begin{gathered} =\frac{a\pm \left(a-12\right)}{4} \\ =\frac{2a-12}{4}or3 \end{gathered}$$

                    $=\frac{a-6}{2}$

Now $-1\le \frac{a-6}{2}\le 1-2\le a-6\le 2$

$\Rightarrow 4\le a\le 8$