Question

Sin⁡(2x-π)=1/2,0≤x≤2π

Answer 1

Dengan demikian,
sin(2x-π) = 1/2
sin(2x-π) = sin(π/6)
Sehingga,
Solusi 1.
2x-π = π/6 + k.2π
2x = 7π/6 + k.2π
x = 7π/12 + kπ
Dengan k = 0, dan k = 1
Didapat:
x = {7π/12, 19π/12}

Solusi 2.
2x-π = (π-π/6) + k.2π
2x-π = 5π/6 + k.2π
2x = 11π/6 + k.2π
x = 11π/12 + kπ
Dengan k = 0 dan k = 1,
Didapat:
x = {11π/12, 23π/12}

HP = {x | 7π/12, 11π/12, 19π/12, 23π/12}