Question 767214
Solve the following equation exactly for 0≤θ≤2Π with θ in radians. 
2sin2(θ) - sin(θ) - 1 = 0
sin(θ)(2sin(θ)-1)=0
..
sin(θ)=0
θ=0, π
..
2sin(θ)-1=0
sin(θ)=1/2
θ=π/6, 5π/6