1708$\boldsymbol{Ans:}$\Rightarrow$ The Equation $(cos p -1)x^2+(cosp)x+sinp=0$ has Real roots,\\\\ Then its $D>0.$ means $cos^2p-4sinp(cosp-1)>0$\\\\ $\underbrace{(cosp)^2-2.cosp.2sinp+(2sinp)^2}+4sinp(1-sinp)>0$\\\\ $(cosp-2sinp)^2+4sinp.(1-sinp)>0$,\\\\ So Here $(cosp-2sinp)^2\geq0$ and $(1-sinp)$ is always $\geq0$ in $p\epsilon R$\\\\ and $sinp>0$ in $p\epsilon(0,\pi)$\\\\ So $\boxed{p\epsilon (0,\pi)}$