Решение
Cos2x=sinx+1
1 - 2sin²x - sinx = 1
2sin²x + sinx = 0
sinx(2sinx + 1) = 0
1) sinx = 0
x = πk, k ∈Z
2) 2sinx + 1 = 0
2sinx = - 1
sinx = - 1/2
x = (-1)^n*arcsin(-1/2) + πn, n ∈ Z
но arcsin (- 1/2) = - arcsin (1/2) = - π/6, тогда
х= (-1)ⁿ⁺¹ π/6 +πn, n∈Z