6t^2-t-2=0
D=1+48=49
t1=-1/2 ИЛИ t2=2/3
cosx=-1/2
± (2π/3)+2πk, k∈Z
ИЛИ
cosx=2/3
x=±arccos (2/3)+2πn,n∈Z
О т в е т.
а) ± (2π/3)+2πk, ±arccos (2/3)+2πn, k, n ∈ Z
б) x1 = (2π/3)-6π= -16π/3
x2 = (-2π/3)-4π=-14π/3
x3=- arccos(2/3)-4π