Recall the half-angle identity for sine,
sin²(x/2) = (1 - cos(x))/2
Then the given equation is identical to
sin²(θ) = 1 - cos(θ)
Also recall the Pythagorean identity,
sin²(θ) + cos²(θ) = 1
Then we rewrite the equation as
1 - cos²(θ) = 1 - cos(θ)
Factoring the left side, we have
(1 - cos(θ)) (1 + cos(θ)) = 1 - cos(θ)
and so
(1 - cos(θ)) (1 + cos(θ)) - (1 - cos(θ)) = 0
and we factor this further as
(1 - cos(θ)) (1 + cos(θ) - 1) = 0
which gives
cos(θ) (1 - cos(θ)) = 0
Then either
cos(θ) = 0 or 1 - cos(θ) = 0
cos(θ) = 0 or cos(θ) = 1
[θ = arccos(0) + 2nπ or θ = -arccos(0) + 2nπ]
… or [θ = arccos(1) + 2nπ or θ = -arccos(1) + 2nπ]
(where n is any integer)
[θ = π/2 + 2nπ or θ = -π/2 + 2nπ] or [θ = 0 + 2nπ]
In the interval 0 ≤ θ < 2π, we get three solutions:
• first solution set with n = 0 ⇒ θ = π/2
• second solution set with n = 1 ⇒ θ = 3π/2
• third solution set with n = 0 ⇒ θ = 0
So, the first choice is correct.