Solve finding all solutions in [0,2pi]?

Question

2sin^2+sin theta=1
2sin^2+sin theta-1=0
write an equation in standard form
factor
use zero-product principle

Answer 1

2 sin^2 θ + sin θ = 1
<=> 2 sin^2 θ + sin θ - 1 = 0
<=> (2 sin θ - 1) (sin θ + 1) = 0
<=> 2 sin θ = 1 or sin θ = -1
<=> sin θ = 1/2 or -1
For θ ∈ [0, 2π] the solutions are
θ = π/6, 5π/6, 3π/2.

Answer 2

First say x=sin(theta)
so you have

2x^2 + x - 1 = 0
(2x-1)(x+1)=0
x = 1/2 or -1
so sin(theta) = 1/2
and sin(theta) = -1

answers are Pi/6, 5Pi/6, and 3Pi/2