Решите неравенство: 6 cos^2 x + cos x -1 = 0

6 cos^2 x + cos x -1 = 0
cosx = t, t∈[-1;1]
6t^1 + t - 1 = 0
D = 1-4*6*(-1) = 25
t1 = (-1+5)/12 = 4/12 = 1/3
t2 = (-1-5)/12 = -6/12 = -1/2
cosx = 1/3
x = ± arrcos1/3 + 2πk, k∈Z
cosx = -1/2
x = ± arrcos(-1/2) + 2πn, n∈Z
x = ± (π - π/3) + 2πn, n∈Z
x = ± (2π/3) + 2πn, n∈Z

Cosx=a
6a²+a-1=0,|a|≤1
D=1+24=25
a1=(-1-5)/12=-1/2⇒cosx=-1/2⇒x=+-2π/3+2πk,k∈z
a2=(-1+5)/12=1/3⇒cosx=1/3⇒x=+-arccos1/3+2⇒k,k∈z