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