Sinx + sin2x = cosx + 2 cos в квадрате x

Sinx + sin2x = cosx + 2cos²x

sinx - cosx + 2sinxcosx - 2cos²x = 0

(sinx - cosx) + 2cosx (sinx - cosx) = 0

(sinx - cosx) (1 + 2cosx) = 0

1) sinx - cosx = 0

sinx = cosx |:cosx

tgx = 1

x = π/4 + πn, n ∈ Z

2) 1 + 2cosx = 0

cosx = - 1/2

x = ±2π/3 + 2πk, k ∈ Z

Ответ: x = π/4 + πn, n ∈ Z; ±2π/3 + 2πk, k ∈ Z.