The equation \((\operatorname{cosp}-1) x^2+(\operatorname{cosp}) x+\sin \mathrm{p}=0\) in the variable \(x\), has real roots. Then p can take any value in the interval

Given equation is
\((\cos \mathrm{p}-1) x^2+(\cos \mathrm{p}) x+\sin \mathrm{p}=0\)
Comparing with \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\), we get
\(a=\cos p-1, b=\cos p, c=\sin p\)
It has real roots.
\(\therefore b^2-4 a c \geq 0\)
\(\Rightarrow \cos ^2 \mathrm{p}-4(\cos \mathrm{p}-1)(\sin \mathrm{p}) \geq 0 \)
\( \Rightarrow \cos ^2 \mathrm{p}-4 \sin \mathrm{p} \cos \mathrm{p}+4 \sin \mathrm{p} \geq 0 \)
\( \Rightarrow \cos ^2 \mathrm{p}-4 \sin \mathrm{p} \cos \mathrm{p}+4 \sin ^2 \mathrm{p} \) \( +4 \sin \mathrm{p}-4 \sin ^2 \mathrm{p} \geq 0\)
\(\Rightarrow(\cos p-2 \sin p)^2+4 \sin p(1-\sin p) \geq 0 \)
\( \therefore (\cos p-2 \sin p) \text { is always positive } \)
\( \therefore 1-\sin p \geq 0 \text { for all values of } p, p \in(0, \pi)\)