1-cos2x=2sin нужно найти х

1 - cos2x = 2 sinx

[ cos2a = cos^2a + sin^2a]

1 - cos^2x + sin^2x = 2sinx

sin^2x + sin^2x = 2sinx

2 sin^2x + 2 sinx = 0

2 sinx (sinx + 1) = 0 | : 2

sinx (sinx + 1) = 0

sinx = 0

x1 = Пn, n € Z

sinx + 1 = 0

sinx = - 1

x2 = - П/2 + 2 Пk, k € Z