I assume the up arrows are supposed to indicate exponents, so that the equation is
sin²(θ) = 2 sin²(θ/2)
Recall the half-angle identity for sine,
sin²(θ/2) = (1 - cos(θ))/2,
as well as the Pythagorean identity,
sin²(θ) + cos²(θ) = 1
Rewrite the equation in terms of cosine and solve:
1 - cos²(θ) = 1 - cos(θ)
cos²(θ) - cos(θ) = 0
cos(θ) (cos(θ) - 1) = 0
cos(θ) = 0 or cos(θ) - 1 = 0
cos(θ) = 0 or cos(θ) = 1
[θ = arccos(0) + 2nπ or θ = arccos(0) - π + 2nπ] or
… … … [θ = arccos(1) + 2nπ]
(where n is any integer)
[θ = π/2 + 2nπ or θ = -π/2 + 2nπ] or [θ = 2nπ]
In the interval 0 ≤ θ < 2π, we get the solutions θ = 0, π/2, and 3π/2.
(That is, for n = 0 in the first and third solution families, and n = 1 in the second family.)