Explanation

(c)
Given equation is $(\cos{p - 1)x^{2} + (\cos{p)x + \sin p = 0}}$
Since, roots are real, its discriminant, $D \geq 0$
$\therefore\cos^{2}{p - 4(\cos{p - 1)\sin{p \geq 0}}}$
$\Rightarrow \cos^{2}{p - 4\cos{p\sin{p + 4\sin p}}} \geq 0$
$\Rightarrow \left( \cos{p - 2\sin p} \right)^{2} - 4\sin^{2}{p + 4\sin{p \geq 0}}$
$\Rightarrow \left( \cos{p - 2\sin p} \right)^{2} + 4\sin{p(1 - \sin{p) \geq 0}}$ …..(i)
Now, $(1 - \sin{p) \geq 0}$ for all real $p$ and $\sin{p > 0}$ for $0 < p < \pi$. Therefore, $4\sin{p(1 - \sin{p) \geq 0}}$ when $0 < p < \pi$ or $p \in (0,\pi)$.