Решите уравнение а) cosx=-1 б) 2sin^2x-sinx-1=0 в) cos^2x-V3sinxcosx=0

A) cosx = - 1

x = π + 2πk, k€Z

б) 2sin²x - sinx - 1 = 0

Пусть t = sinx, t€[-1; 1].

2t² - t - 1 = 0

D = 1 + 2•4 = 9 = 3²

t1 = (1 + 3) / 4 = 1

t2 = (1 - 3) / 4 = - 1/2

Обратная замена:

sinx = 1

x = π/2 + 2πk, k€Z

sinx = - 1/2

x = (-1) ⁿ+¹π/6 + πk, k€Z.

в) cos²x - √3sinxcosx = 0

cosx (cosx - √3sinx) = 0

cosx = 0

x = π/2 + πk, k€Z

cosx = √3sinx

1 = √3tgx

tgx = √3/3

x = π/6 + πk, k€Z.

Ответ: x = π/2 + πk, k€Z; x = π/6 πk, k€Z.