The equation $k\sin x+\cos 2x=2k-7$ possess a solution, if

Solution:

We have,

$\begin{array}{l}k\sin x+\left(1-2{\sin }^{2}x\right)=2k-7\\ \Rightarrow 2{\sin }^{2}x-k\sin x+2(k-4)=0\\ \Rightarrow \sin x=\frac{k\pm \sqrt{k^2-16k+64}}{4}=\frac{k\pm (k-8)}{4}=\frac{1}{2}(k-4),2\\ \Rightarrow \sin x=\frac{1}{2}(k-4)[\because \sin x\ne 2]\end{array}$

Now, $-1\le \sin x\le 1\Rightarrow -1\le \frac{k-4}{2}\le 1\Rightarrow 2\le k\le 6$