8 sin x + 4 cos ^2 x = 7

Помогите!

8sinx + 4 (1 - sin^2x) - 7 = 0

8sinx + 4 - 4 sin^2x - 7 = 0

- 4sin^2x + 8sinx - 3 = 0 / : (-1)

4sin^2x - 8sinx + 3 = 0

Пусть sinx = t, | t | ≤ 1

D = 64 - 48 = 16 = 4^2

t1 = (8 + 4) / 8 = 12/8 = 1,5 ∉ | t | ≤ 1

t2 = (8 - 4) / 8 = 1/2

Обратная замена

sinx = 1/2

x = pi/6 + 2pik, k ∈ Z

x = 5pi/6 + 2pik, k ∈ Z