Explanation

(a)
$\because\ \sin^{5}{x - \cos^{5}{x = \frac{\sin{x - \cos x}}{\sin{x\cos x}}}}$
$\Rightarrow \ \sin{x\cos{x\left\lbrack \frac{\sin^{5}{x - \cos^{5}x}}{\sin{x - \cos x}} \right\rbrack = 1}}$
$\Rightarrow \ \frac{1}{2}\sin{2x\lbrack\sin^{4}{x + \sin^{3}{x\cos{x + \sin^{2}{x\cos^{2}{x + \sin{x\cos^{2}{x + \cos^{4}{x\rbrack = 1}}}}}}}}}$
$\Rightarrow \ \sin{2x\lbrack}\left( \sin^{2}{x + \cos^{2}x} \right)^{2} - 2\sin^{2}{\ x\cos^{2}x}$
$+ \sin{x\cos{x(\sin^{2}{x + cos^{2}x) + \sin^{2}{x + \ }}}}\cos^{2}{x\rbrack = 2}$
$\Rightarrow \sin{2x\lbrack 1 - \sin^{2}{x\cos^{2}{x + \sin{x\cos{x\rbrack = 2}}}}}$
$\Rightarrow \sin^{3}{2x - 2\sin^{2}{2x - 4\sin{2x + 8 = 0}}}$
$\Rightarrow ({\sin{2x - 2)\ }}^{2}(\sin{2x + 2) = 0}$
$\Rightarrow \sin{2x = \pm 2}$, which is not possible for any $x$