До завтра!помогите,буду благодарна 1. Решите уравнение а) ctg = 3 б) sin ( - 2x) = 0
A) √3ctg(x/2) = 3 ctg(x/2) = 3/√3 ctg(x/2) = √3 x/2 = π/6 + πk, k€Z x = π/12 + πk/2, k€Z. b) sin(π/6 - 2x) = 0 -sin(2x - π/6) = 0 sin(2x - π/6) = 0 -π/6 + 2x = πk, k€Z 2x = π/6 + πk, k€Z x = π/12 + πk/2, k€Z.