3 tan² beta=1,π/2<beta<π

Jawab

3 tan² β  = 1
tan² β = 1/3
tan β = +_√(1/3)
tan β = 1/3 √3 atau tan β = – 1/3 √3

β pada interval π/2  ≤ β ≤ π → β dikuadran II→ tan β < 0
tan β  = – 1/3 √3 = – tan 30 = tan(180 -30) = tan 150
β = 150 = 150/180 π
β  = 5/6 π