Question

2sin(2x+pi/6)+1=корень3×sin2x+cosx

Answer 1

2Sin(2x + π/6) + 1 = √3Sin2x + Cosx

2(Sin2xCosπ/6 + Cos2xSinπ/6) + 1 = √3Sin2x + Cosx

2(Sin2x * √3/2 + Cos2x * 1/2) + 1 = √3Sin2x + Cosx

√3Sin2x + Cos2x + 1 = √3Sin2x + Cosx

Cos2x - Cosx + 1 = 0

2Cos²x - 1 - Cosx + 1 = 0

2Cos²x - Cosx = 0

Cosx(2Cosx - 1) = 0

1) Cosx = 0

x = π/2 + πn , n ∈ Z

2) 2Cosx - 1 = 0

2Cosx = 1

Cosx = 1/2

x= ± arcCos1/2 + 2πn , n ∈ Z

x = ± π/3 + 2πn , n ∈ Z