The equation (cosp−1)x2+(cosp)x+sinp=0  has real roots, then ‘p’ can take any value in the interval is

The given equation is (cosp−1)x2+(cosp)x+sinp=0 And the roots are real so Δ≥0 a=cosp−1,b=cosp,c=sinp ⇒cos2p−4(cosp−1)sinp≥0      (∵ b2−4ac≥0)  add & subtract 4sin2p  ⇒cos2p+4sin2p−4cospsinp+4sinp−4sin2p≥0 ⇒(cosp−2sinp)2+4sinp(1−sinp)≥0    ⇒4sinp(1−sinp)>0  ⇒sinp>0⇒p∈(0,π) ∴  the interval is(0,π)