Question

Given that 0 ° < θ < 360 ° and that one solution for θ is 30°, find the other two possible values for θ

Answer 1

Any value between 0 and 360 that isn't 30

Answer 2

2sin²θ + sin θ− 1=0
→(sinθ+1)(2sin θ− 1)=0
Hence either (sinθ+1) = 0 i.e sin θ =− 1 i.e θ=− 270°
or (2sin θ− 1)=0 i.e sinθ=1/2 i.e θ=30° or 150°
Hence possible values of θ are θ =− 270° ; 30° ; 150