The equation  cosp−1x2+cospx+sinp=0 in the variable x has real roots. Then p can take any value in the interval

Since the equation cosp−1x2+cospx+sinp=0 has real roots ⇒b2-4ac≥0⇒cos2p−4sinpcosp−1≥0⇒since  cos2p≥0  and  cosp−1≤0,One of the possibilities for the inequality to hold good is  sinp≥0⇒p∈0,π