3(2cos^2(x) - 1) - 5cosx = 1
6cos^2(x) - 5cosx - 4 = 0
Замена: cosx = t, t∈[-1;1]
6t^2 - 5t - 4 = 0
D = 25 + 4*6*4 = 121
t1 = (5 - 11)/12 = -6/12 = -1/2
t2 = (5+11)/12 = 16/12 > 1 - посторонний корень
cosx = -1/2
x = 2π/3 + 2πk, k∈Z
x = 4π/3 + 2πk, k∈Z
или можно записать по-другому:
x = +-2π/3 + 2πk, k∈Z