Рeшите: ctg 2x + 4 = 0 2sin 2x + 1 = 0 tg 3x - 1 = 0 √2 cosx - 1 = 0
сейчас решу
Ctg2x=-4 2x=arcctg(-4)+πn,n∈z x=1/2*arcctg(-4)+πn/2,n∈z sin2x=-1/2 2x=(-1)^(n+1)*π/6+πn,n∈z x=(-1)^(n+1)*π/12+πn/2,n∈z tg3x=1 3x=π/4+πn,n∈z x=π/12+πn/3,n∈z cosx=1/√2 x=+-π/4+2πn,n∈z