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.