the eq: (cos p-1)x^2+(cos p)x + sin p=0,where x is a variable ,has real roots.find the interval of p?

Question

1)0,2pi
2)0,pi

Answer 1

Hint: It's a quadratic equation in x. Therefor it will have real roots whenever √((cos²(p)-4cos(p-1)sin(p)) is real, or when (cos²(p) - 4cos(p-1)sin(p) > 0.


Doug

Answer 2

The condition for the equation in x to have real roots is that
the discriminant is non negative

cos^2 p - 4(sin p)(cos p - 1) >= 0

If you had equality then the solution for p is very complicated.
I used Mathematica and there were two real roots expressed in terms of arccosines of radicals.

I would recomend the use of a graphing calculator
Graph the function cos^2 p - 4(sin p)(cos p - 1)
and see what intervals for p it is above or on the p axis