sinx + sin2x = cosx + 2 cos в квадрате x

sinx + sin2x = cosx + 2cos²xsinx - cosx + 2sinxcosx - 2cos²x = 0(sinx - cosx) + 2cosx(sinx - cosx) = 0(sinx - cosx)(1 + 2cosx) = 01) sinx - cosx = 0sinx = cosx         |:cosxtgx = 1x = π/4 + πn, n ∈ Z2) 1 + 2cosx = 0cosx = -1/2x = ±2π/3 + 2πk, k ∈ ZОтвет: x = π/4 + πn, n ∈ Z; ±2π/3 + 2πk, k ∈ Z.