Question

6sin в квадрате x-5cosx+5=0

Answer 1

sin^2 x=1-cos^2 x

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

6-6cos^2 x - 5cos x + 5=0 

 -6cos^2x-5cos x + 11=0

6cos^2 x + 5 cos x -11=0 

cos x=y

6y^2+5y-11=0

D= 289

y1=(-5+17)/12=1

y2= (-5-17)/12=-11/6 - не подходит, т.к. -1<=сos x<=1

cos x=1

x=0