The eq. (cosp - 1)x2 +cospx +sinp = 0 where x is a variable, has real roots. then the interval of p is.........................

Dear piyush,

The given equation is (cosp - 1)x² + cos px + (1 - sin p) = 0,

Discriminant D = b² - 4ac ≥ 0.  given x has real roots
 

=> cos² p - 4(1 - sin p)(cos p - 1) ≥ 0
 
cos² p is always positive and (1 - sin p) is also positive. because (sin p <1)

(cos p - 1) is always negative.(cos p <1)

Hence the given expression is positive for all the values of p.And p can be in any of the quadrants.

 =>   p ∈ (- ∞ to ∞)

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