[" The equation "(cos p-1)x^(2)+(cos p)x+sin p=0],[" has real roots in "x" ,then set of values of "p'" is: "]

The equation (cos p-1)x^(2)+cos p*x+sin p=0 in x has real roots.Then the set of values of p is (a) [0,2pi](b)[-pi,0](c)[-(pi)/(2),(pi)/(2)](d)[0,pi]

The equation (cos p-1)x^(2)+cos px+sin p=0 where x is a variable,has real roots.then the interval of p may be any one of the following:

The equation (cos p-1)x^(2)+(cos p)x+sin p=0 in the variable x has real roots.The p can take any value in the interval (a) ( 0,2 pi)( b) (-pi,0) (c) (-(pi)/(2),(pi)/(2))( d )(0,pi)

If the roots of the given equation (cos p-1)x^(2)+(cos p)x+sin p=0 are real if

If the equation (cos p - 1)x^(2) + cos p x + sin p =0 in the varable x has real roots then p can taken any value in the interval

The equation (cos p-1) x^(2) + cos p*x + sin p = 0 where x is a variable, has real roots. Then the interval of possible values of p is

The equation (cosp-1)x^2+(cos p)x+sin p=0 in the variable x has real roots. The p can take any value in the interval (a) (0,2pi) (b) (-pi,0) (c) (-pi/2,pi/2) (d) (0,pi)

The equation (cosp-1)x^2+(cos p)x+sin p=0 in the variable x has real roots. The p can take any value in the interval (a) (0,2pi) (b) (-pi,0) (c) (-pi/2,pi/2) (d) (0,pi)

The equation (cosp-1)x^2+(cos p)x+sin p=0 in the variable x has real roots. The p can take any value in the interval (a) (0,2pi) (b) (-pi,0) (c) (-pi/2,pi/2) (d) (0,pi)