решите уравнение 2cos^2x+sin x +1 = 0

1)2cos^2 x-sin x-1=02(1-sin^2 x)-sin x-1=02-2sin^2 x-sin x-1=02sin^2 x+sin x-1=0sin x=a2a^2+a-1=0D=1+8=9a1=(-1+3)/4=0,5a2=(-1-3)/4=-11) sin x=0,5x=(-1)^n*pi/6+pi*n, n E Z2) sin x=-1x=-pi/2+2*pi*n, n E Z