Question

6 cos(в квадрате)x+cos x-1=0

Answer 1

cosx=t

6t²+t-1=0

D=1+4=5

√D=√5

t₁=cosx₁=(-1+√5)/12   ===> x₁=±arccos ((-1+√5)/12) 

t₂=cosx₂=(-1-√5)/12   ===> x₂=±arccos ( (-1-√5)/12 )