If the roots of the given equation (cos p-1)x2+(cos p)x+sin p=0 are real, then
p∈(-π, 0)
p∈(-π2, π2)
p∈(0,π)
p∈(0,2π)
Given equation (cos p-1)x2+(cos p)x+sin p=0
Its discriminant D≥0 since roots are real
⇒cos2 p-4 cos p sin p+4 sin p≥0 ⇒cos2 p-4 cos p sin p+4 sin p≥0 ⇒(cos p-2 sin p)2-4 sin2 p+4 sin p≥0 ⇒(cos p-2 sin p)2+4 sin p(1-sin p)≥0 ....(i)
Now (1-sin p)≥0 for all real p, sin p>0 for
0<p<π. Therefore 4 sin p(1-sin p)≥0 when
0<p<π or p∈(0,π)