Explanation

(a)
Given, $\cos{2x + k\sin{x = 2k - 7}}$
$\Rightarrow \ \ 1 - 2\sin^{2}{x + k\sin{x = 2k - 7}}$
$\Rightarrow \ 2\sin^{2}{x - k\sin{x + 2k - 8 = 0}}$
$\Rightarrow \ \sin{x = \frac{+ k \pm \sqrt{k^{2} - 8(2k - 8)}}{4}}$
$\Rightarrow \ \sin{x = \frac{k \pm (k - 8)}{4}}$
$\Rightarrow \ \sin{x = \frac{k - 4}{2}}$
$(\because\ for\ ' -^{'}sign\sin{x = 2},\ which\ is\ not\ possible)$
$\because\ - 1 \leq \sin x \leq 1\ \ \Rightarrow \ - 1 \leq \frac{k - 4}{2} \leq \ 1$
$\Rightarrow \ - 2 \leq k - 4 \leq 2\ \ \ \ \Rightarrow \ \ 2 \leq k \leq 6$