6 cos в квадрате + 5cos+1=0

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6 cos в квадрате + 5cos+1=0

6 cos² x + 5cosx+1=0

6t ²+5t+1=0

t1=[-5-√ (25-24) ]/12=-1/2 cosx = - 1/2 x1=π-π/3+2πn=2π/3+2πn, n∈Z

x2 = - (π-π/3) + 2πn=-2π/3+2πn, n∈Z

t2=[-5+√ (25-24) ]/12=-1/3 cosx = - 1/2 x1=π-arccos (1/3) + 2πn, n∈Z

x2 = - (π-arccos (1/3)) + 2πn, n∈Z