6sin^2x-5cosx-5=0 Прошу решите)

6 (1-cos^2x) - 5cosx-5=0

6-6cos^2x-5cosx-5=0

6cos^2x+5cosx+5-6=0

6cos^2x+5cosx-1=0

Cosx=t

6t^2+5t-1=0

D=b^2-4ac

D=25+24=49

t1 = (-5+7) / 12=1/6

t2 = (-5-7) / 12=-1

t=cosx

1) cosx=-1

X=П+2 ПК; К€Z

2) cosx=1/6

x=+-arccos1/6+2 ПN; N€Z