JEE Advance - Mathematics (1990 - No. 9)
The equation $$\left( {\cos p - 1} \right){x^2} + \left( {\cos p} \right)x + \sin p = 0\,$$
In the variable x, has real roots. Then p can take any value in the interval
$$\left( {0,2\pi } \right)\,$$
$$\left( { - \pi ,0} \right)\,\,\,$$
$$\left[ { - {\pi \over 2},{\pi \over 2}} \right]\,$$
$$\left( {0,\pi } \right)$$