12cos²x + 10sinx = 4 ║: 2
6cos²x + 5sinx - 2 = 0
6•(1 - sin²x) + 5sinx - 2 = 0
6 - 6sin²x + 5sinx - 2 = 0
- 6sin²x + 5sinx + 4 = 0
6sin²x - 5sinx - 4 = 0
Пусть sinx = a , a ∈ [ - 1 ; 1 ] , тогда
6a² - 5a - 4 = 0
D = (-5)² - 4•6•(-4) = 25 + 96 = 121 = 11²
a₁ = (5 - 11)/12 = - 6/12 = - 1/2
a₂ = (5 + 11)/12 = 16/12 = 4/3 ∉ [ - 1 ; 1 ]
a = - 1/2 ⇔ sinx = - 1/2
[ x = - (п/6) + 2пn
[ x = - (5п/6) + 2пn , n ∈ Z
ОТВЕТ: - (п/6) + 2пn ; - (5п/6) + 2пn , n ∈ Z