Question

Solve the following equation for θθ, where 0 ≤ θ < 2π radians. This time, give exact

answer(s) for θθ.

2 sin^2 θ+ sin θ − 1 = 0

Answer 1

Explanation:

2sin²θ+sin θ-1=0

2sin²θ+2sinθ-sin θ-1=0

2sin θ(sin θ+1)-1(sin θ+1)=0

(sin θ+1)(2sin θ-1)=0

either sin θ+1=0

sin θ=-1=sin (3π/2)

θ=3π/2

or

2 sinθ-1=0

sin θ=1/2=sin (π/6),sin (π-π/6)

θ=π/6,5π/6