p can take any value in the interval? 
If the equation $(\cos(p)-1)x^2+\cos(p)x+\sin(p)=0$ in the variable $x$ has real
  roots, then $p$ can take any value in what interval?

I applied the discriminant $D>0$. I get $\cos^2( p)-4(\cos(p)-1)\sin(p)>0$. Hence,$\left(\frac{1+\cos(2p)}{2}\right)^2-4(\cos(p)-1)\sin(p)>0$. I do get a $\sin(2p)$ and $\cos(2p)$ but the $\sin(p)$ term is apparently making it complicated. How to solve further?

Just put the product of the roots to be negative $i.e.$ 
$\frac{sin(p)}{cos(p)-1}<0$ which will give $tan(\frac{p}{2})>0$. So final solution for the $p$ will be 
$$p\in (2n\pi,(2n+1)\pi)\cup((2n+2)\pi,(2n+3)\pi)$$
Hope this will be helpful !!!
