Question

1 + sin^2x + cosx = 0

Answer 1

1 + sin²x + cosx = 0
sin²x = 1-cos²x
1 + 1-cos²x + cosx = 0
cos²x - cosx - 2 = 0   cosx = t , t ∈ [-1;1]
t² - t - 2 = 0
t1 = (1 - √9) / (2∙1) = -1
t2 = (1 + √9) / (2∙1) = 2 не подходит 2 ∉  [-1;1]
cosx = -1
x = π + 2πk, k ∈ Z