Question

For 0 less than or equal to theta less than 2(pi), what are thebsolutions to sin↑2(theta)=2(sin↑2)(theta/2)? ​

Answer 1

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.)