Question

2cos la2 x +3sinx=0 rezolvați inr ecuatia​

Answer 1

Rezolvare :

2cos² x-3sin x = 0

2(1-sin² x ) + 3sin x = 0

2-2sin² x + 3sin x = 0

- 2sin²x + 3sin x + 2 = 0

2sin²x-3sin x - 2 = 0 => 2sin² x - 4sin x + sin x - 2 = 0 =>

2sin x (sin x-2)+1(sin x-2) = 0

(sin x - 2 ) (2sin x + 1 ) = 0 => sin x = -1/2 sau sin x = -1

x ∈ [ - π/2 , + π/2 ] ; x= - π/6 / x= -π/2