Explanation

(a)
$\because\ \sin{2x + \cos{4x = 2}}$
It is possible only when
$\sin{2x = 1\ and\cos{4x = 1}}$
$\Rightarrow 2x = 2n\pi + \frac{\pi}{2}\ and\ 2x = 2m\pi$
$\therefore\ x = n\pi + \frac{\pi}{4}\ and\ x = m\pi,\ n \in I$
Then, solution$= \left( n\pi + \frac{\pi}{4},\ n \in I \right) \cap (m\pi,\ m \in I) = \phi$