Explanation

(a)
Given, $\sin^{4}{x + \cos^{4}{x = \sin{x \bullet \cos{x\ }}}}$
$\Rightarrow \ \left( \sin^{2}{x + \cos^{2}x} \right)^{2} - 2\sin^{2}{x \bullet \cos^{2}{x = \sin{x \bullet \cos x}}}$
$\Rightarrow \ 1 - \frac{\sin^{2}{2x}}{2} = \frac{\sin{2x}}{2}$
$\Rightarrow \ \sin^{2}{2x + \sin{2x - 2 = 0}}$
$\Rightarrow \ (\sin{2x + 2)(\sin{2x - 1) = 0}}$
$\Rightarrow \ \sin{2x = 1\ \ \ (\because\sin{2x \geq - 1)}}$
$\therefore 2x = (4n + 1)\frac{\pi}{2}$
$\Rightarrow \ x = (4n + 1)\frac{\pi}{4}$
$\Rightarrow \ x = \frac{\pi}{4},\ \frac{5\pi}{4}$
Hence, two solutions exist