РЕШИТЕ УРАВНЕНИЯ ПОЖАЛУЙСТА а) cos ( x/2 + π/4 ) + 1 = 0 б) 2cos π/6 * tgx * ctgx = 0 в) cos (-x) = √3/2

Решение а) cos ( x/2 + π/4 ) + 1 = 0 cos ( x/2 + π/4 )  = - 1 x/2 + π/4 = π + 2πk, k∈Z x/2 = π - π/4 + 2πk, k∈Z x/2 = 3π/4 + 2πk, k∈Z x = 3π/2 + 4πk, k∈Z б) 2cos π/6 * tgx * ctgx = 0 2*(√3/2) * tgx * ctgx = 0  tgx * ctgx = 0 tgx  = 0 x = πn, n∈Z ctgx = 0 x = π/2 + πk, k∈Z в) cos (-x) = √3/2 cosx = √3/2 x = (+ -)arccos(√3/2) + 2πm, m∈Z x = (+ -)*(π/6) + 2πm, m∈Z