Question

The solution of equation cos'2q+sinq+1 =0 lies
in the interval​

Answer 1

Step-by-step explanation:

cos^2 x + sin x + 1 = 0 

=>1 - sin^2 x + sin x +1 = 0 

=>- sin^2 x + sin x +2 = 0 <=> sin^2 x - sin x -2 = 0 

Let z = sin x => z^2 - z -2 = 0 

=> (z + 1)(z - 2) = 0 

=> z = 2 => sin x = 2 (impossible => rejected) 

or 

=> z = -1 => sin x = -1 => sin x = sin (-90 deg) 

=> x = (180n - (-1)^n * 90) deg, n e Z (i.e. n is an integer) 

Therefore, x lies in the interval [Neg Infinitif, Pos Infinitif] & has the form above!