Решите уравнения.5sin2х+6cosx-6=0

[latex]5sin ^{2} x+6cosx-6=0[/latex] ОДЗ:х∈R [latex]5*(1-cos ^{2} x)-6cosx-6=0[/latex] [latex]5-5cos ^{2} x-6cosx-6=0[/latex] [latex]-5cos ^{2} x-6cosx-1=0[/latex] cosx=t -5t²-6t-1=0 D=16>0(2рдк) t1= -1 t2= -1/5 cosx= -1     или      cosx= -1/5 x= П+2Пn; n∈Z      x= П-arccos1/5+2Пn;n ∈Z