Математика. Решите пожалуйста!!!
2sinx*cosx - sinx - (2cosx - 1) = 0 sinx(2cosx - 1) - (2cosx - 1) = 0 (2cosx - 1)(sinx - 1) = 0 2cosx = 1 cosx = 1/2 x₁ = +/-π/3 + 2πk, k∈Z sinx = 1 x₂ = π/2 + 2πn, n∈Z Ответ: x₁ = +/-π/3 + 2πk, k∈Z; x₂ = π/2 + 2πn, n∈Z