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.