Вычислите cos2x+sinx=sin3x

1 - 2sin²x + sinx = 3sinx - 4sin³x
4sin³x - 2sin²x - 2sinx + 1 = 0
2sin²x(2sinx - 1) - (2sinx - 1) = 0
(2sin²x - 1)(2sinx - 1) = 0
2sin²x = 1
sin²x = 1/2
sinx = ±1/2
x = (-1)ⁿ+¹π/6 + πn, n ∈ Z
x = (-1)ⁿπ/6 + πn, n ∈ Z.

2sinx - 1 = 0
sinx = 1/2
x = (-1)ⁿπ/6 + πn, n ∈ Z.

Ответ: х = (-1)ⁿπ/6 + πn, n ∈ Z; x = (-1)ⁿ+¹π/6 + πn, n ∈ Z.