2cos^2 x=1+sin x

Прошу быстрее

2cos²x = 1 + sinx

2 (1 - sin²x) = 1 + sinx

2 - 2sin²x = 1 + sinx

2sin²x + sinx - 1 = 0

Пусть sinx = t, причем t ∈ [-1; 1]

2t² + t - 1 = 0

D = 1 + 8 = 9

t = (- 1 + 3) / 4 = 1/2;

t = (- 1 - 3) / 4 = - 1;

1) sinx = 1/2

x = (-1) ^k*arcsin (1/2) + πk

x = (-1) ^k*π/6 + πk, k ∈ Z

2) sinx = - 1

x = - π/2 + 2πk, k ∈ Z

Ответ:

(-1) ^k * π/6 + πk;

-π/2 + 2πk, k ∈ Z