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.