The equation $(\cos p-1)x^2+\left(\cos p\right)x+\sin p=0$  has real roots, then ‘p’ can take any value in the interval is

The given equation is $(\cos p-1)x^2+\left(\cos p\right)x+\sin p=0$

And the roots are real so $\Delta \ge 0$

$a=\cos p-1,b=\cos p,c=\sin p$

$\Rightarrow {\cos }^{2}p-4\left(\cos p-1\right)\sin p\ge 0$      $\left(\because b^2-4ac\ge 0\right)$

 add & subtract $4{\sin }^{2}p$  

$\Rightarrow {\cos }^{2}p+4{\sin }^{2}p-4\cos p\sin p+4\sin p-4{\sin }^{2}p\ge 0$

$\Rightarrow {\left(\cos p-2\sin p\right)}^2+4\sin p\left(1-\sin p\right)\ge 0$  

 $\Rightarrow 4\sin p\left(1-\sin p\right)>0$

 $\Rightarrow \sin p>0\Rightarrow p\in \left(0,\pi \right)$

$\therefore$  the interval is$\left(0,\pi \right)$