Question

Prove that the general solution of cos θ = cos α is θ = 2nπ ± α, n ∈ Z.

Answer 1

cos θ = cos α

∴ cos θ – cos α = 0

∴ `-2sin((θ + α)/2)*sin((θ - α)/2)` = 0

∴ `sin((θ + α)/2)` = 0 or `sin((θ - α)/2)` = 0

∴ `(θ + α)/2` = nπ or `(θ - α)/2` = nπ, n ∈ Z

∴ θ + α = 2nπ or θ – α = 2nπ, n ∈ Z

∴ θ = 2nπ – α or θ = 2nπ + α, n ∈ Z

∴ θ = 2nπ ± α, n ∈ Z

Hence, the general solution of cos θ = cos α is θ = 2nπ ± α, n ∈ Z.