Explanation

(c)
Given, $\tan{x + \sec{x = 2\cos x}}$
On multiplying by $\cos{x \neq 0}$, we get
$\sin{x + 1 = 2\cos^{2}x}$
$\Rightarrow \ \sin{x + 1 = 2(1 - \sin{x)(1 + \sin{x)}}}$
$\Rightarrow \ (\sin{x + 1)(2\sin{x - 1) = 0}}$
$\Rightarrow \ \sin{x = - 1\ and\ \sin{x = \frac{1}{2}}}$
$\because\ \sin{x \neq - 1\ \ \ (\because\cos{x \neq 0)}}$
$\therefore\ \sin{x = \frac{1}{2}}$
$\Rightarrow \ x = \frac{\pi}{6},\frac{5\pi}{6}$