Cos2x=sinx+1 ...

Решение

Cos2x=sinx+1

1 - 2sin ²x - 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