(A) Given that 2sin^2(θ) + sin(θ) - 1 = 0, find the two values for sin θ
(B) Given that 0° ≤ θ ≤ 360° and that one solution for θ is 30°, find the other two possible values for θ.
A: Let x = sin (theta)
then 2x^2 +x -1 = 0 then use quadratic formula to find x (which is sin(theta))
B: Use your answer from A to determine theta possibilities