Explanation

(b)
Given, $\cos{x + \sin{x = \frac{1}{2}}}$
$\Rightarrow \ \ 1 + \sin{2x = \frac{1}{4}}$
$\Rightarrow \ \frac{2\tan x}{1 + \tan x} = \frac{- 3}{4}$
$\Rightarrow \ \ 8\tan{x = - 3 - 3\tan^{2}x}$
$\Rightarrow \ 3\tan^{2}{x + 8\tan{x + 3 = 0}}$
$\Rightarrow \ \ \tan{x = \frac{- 8 \pm \sqrt{64 - 36}}{6}} = \frac{- 8 \pm 2\sqrt{7}}{6}$
$\Rightarrow \ = - \left( \frac{4 \pm \sqrt{7}}{3} \right)$