Use s for sinθ, 5s^2-16s+3=0 and factor, (s-3)(5s-1)=0 so
either s=3 and NO SOLUTION,
or s=0.2 , θ=0.201 or 2.940 so
θ=0.201+2πk or 2.940+2πk.
5 sin^2 θ − 16 sin θ + 3 = 0
→5 sin² θ − 15 sin θ− sin θ+ 3 = 0
→5 sin θ(sin θ − 3) − 1( sin θ− 3) = 0
→(sin θ − 3)(5 sin θ− 1) = 0
Either (sin θ − 3) = 0 i.e sin θ = 3 which does not have a ion(5 sin θ− 1) = 0
or 5 sin θ− 1 = 0 i.e sin θ = 1/5 θ =sin‾¹(1/5)
In general θ =(2k+1) π − sin‾¹(1/5) and θ =2k π + sin‾¹(1/5) for 0 ≤ θ < 2π
5 sin^2 θ − 16 sin θ + 3=0
5 sin^2 θ − 15 sin θ - sin θ + 3=0
5 sin θ (sin θ-3)-1(sin θ-3) =0
(5 sin θ-1)(sinθ-3)=0
sin θ=1/5,3
