Verified answer

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