Interval in which $(\cos p-1)x^2+\cos p.x+\sin p=0$ has real roots

The equation $$(\cos p-1)x^2+\cos p.x+\sin p=0$$ where $$x$$ is a variable has real roots. Then the interval of $$p$$ may be any of the following: $$(a)\quad(0,2\pi)\quad (b)\quad (-\pi,0)\quad (c)\quad \big(\frac{-\pi}{2},\frac{\pi}{2}\big)\quad (d)\quad(0,\pi)$$

The solution given in my reference is the interval $$(0,\pi)$$.

My Attempt $$\Delta=\cos^2p-4(\cos p-1)\sin p\geq 0\\ \cos^2p-4\sin p\cos p+4\sin p\geq 0\\ \Delta'=16\sin^2p-16\sin p\leq0\implies\sin^2p\leq\sin p\\$$

Similar problem has been asked before Find the range of values of $$p$$ if $$(\cos p−1)x^2+(\cos p)x+\sin p=0$$ has real roots in the variable $$x$$. but it does not address how to prove it analytically.

Adding to what you did, Let us solve the inequation

$$\sin^2 p\le \sin p$$ or

$$\sin p(\sin p-1)\le 0$$

which yields to

$$\sin p\ge 0$$ because $$\; (\sin p-1)\le 0$$.

the answer is $$d) : 0.

Hint :

$$\cos^2p - 4\sin p \cos p + 4\sin p = (\cos p - \sin p)^2 +4\sin p - \sin^2p$$

$$=$$

$$(\cos p - \sin p )^2 + (4-\sin p)\sin p$$

Thus, the consraint yielded for real solutions is translated to :

$$(\cos p - \sin p )^2 + (4-\sin p)\sin p \geq 0$$