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