Решите уравнение 2cos2x+cosx=sin (3 п/2+x) - 2

Sin (3π/2+x) = - cosx

2cos2x+cosx = - cosx-2

2cos2x+2cosx = - 2

2 (cos2x+cosx) = - 2

cos2x+cosx = - 1

cosx = - 1-cos2x

cosx = - (sin²x+cos²x) - (cos²x-sin²x)

cosx = - sin²x-cos²x-cos²x+sin²x

cosx = - 2cos²x

2cos²x+cosx = 0

cosx (2cosx+1) = 0

cosx = 0 2cosx+1 = 0

1) x = π/2+πk, k∈Z

2) cosx = - 1/2

x=2π/3+2πk, k∈Z