determine the solution set for 2sin^2x =1+sinx for 0 ≤ x ≤2π?

2sin²x = 1 + sinx
2sin²x - sinx - 1 = 0
(2sinx + 1)(sinx - 1) = 0
sinx = -1/2 or sinx = 1
x = π + (π/6), 2π - (π/6) or x = π/2
x = π/2, 7π/6, 11π/6

2 sin²x - sin x - 1 = 0
[ 2 sin x + 1 ] [ sin x - 1 ] = 0
sin x = - 1/2 , sin x = 1
x = 7π / 6 , 11π / 6 , π /2

2.sin²(x) = 1 + sin(x)

2.sin²(x) - sin(x) - 1 = 0

2.sin²(x) - [2.sin(x) - sin(x)] - 1 = 0

2.sin²(x) - 2.sin(x) + sin(x) - 1 = 0

[2.sin²(x) - 2.sin(x)] + [sin(x) - 1] = 0

2.sin(x).[sin(x) - 1] + [sin(x) - 1] = 0

[sin(x) - 1].[2.sin(x) + 1] = 0


First case: [sin(x) - 1] = 0 → sin(x) - 1 = 0 → sin(x) = 1

x₁ = π/2


Second case: [2.sin(x) + 1] = 0 → 2.sin(x) + 1 = 0 → 2.sin(x) = - 1 → sin(x) = - 1/2

x₂ = π + (π/6) → x₂ = 7π/6

x₃ = 2π - (π/6) → x₃ = 11π/6