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?
 A: I don't see any other way except layman's approach:
First  of  all, it will be $\ge $ and not $>$
Putting $\cos(p)=x$
$$4(x-1)\sin(p)\le x^2$$$$\implies \sin(p)\ge{x^2\over 4(x-1)}$$
Now since the RHS is $\le 0\therefore$ for all $x$ such that $\sin(\cos^{-1}x)\ge0$, the inequality holds.$\therefore S_1=\{z|z\in \Bbb{R},\sin(\cos^{-1}z)\ge0\}$ 
Now we need to consider only those cases that $\sin(p)\lt0$ 
As both the sides are $-ve$, so squaring will reverse the signs$$\implies \sin^2(p)\le{x^4\over 16(x-1)^2}$$$$\implies 16(x-1)^2(1-x^2)\le x^4$$$$\implies 17x^4-32x^3+32x-16\ge0$$
Check the solution here
The solution set for the above equation will be $(-\infty,-0.992]\cup[0.769,\infty)$
Now since there was the initial condition that $\sin(p)\lt0$ the solution set for the second part will be : $S_2=\{y|y\in(-\infty,-0.992]\cup[0.769,\infty), \sin(\cos^{-1}y)\lt0\}$
The ultimate answer set (the values of $p$) is $\cos^{-1}(S_1)\cup \cos^{-1}(S_2)$
A: 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 !!!
