Question

Find the general solution of the following equation:

cos 4θ = cos 2θ

Answer 1

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

∴ the general solution of cos 4θ = cos 2θ is given by
4θ = 2nπ ± 2θ, n ∈ Z

Taking positive sign, we get
4θ = 2nπ + 2θ, n ∈ Z
∴ 2θ = 2nπ, n ∈ Z
∴ θ = nπ, n ∈ Z

Taking negative sign, we get
4θ = 2nπ – 2θ, n ∈ Z
∴ 6θ = 2nπ, n ∈ Z
∴ θ = `(npi)/(3)`, n ∈ Z

Hence, the required general solution is

θ = `(npi)/(3)`,n ∈ Z or θ = nπ, n ∈ Z.

Alternative Method:

cos 4θ = cos 2θ
∴ cos 4θ – cos 2θ = 0

∴ `-2sin((4θ + 2θ)/2).sin((4θ - 2θ)/2)` = 0

∴  sin 3θ. sin θ = 0

∴ either sin 3θ = 0 or sin θ = 0

The general solution of sin θ = 0 is θ = nπ, n ∈ Z.

∴ the required general solution is given by

3θ = nπ, n ∈ Z or θ = nπ, n ∈ Z

i.e. θ = `(npi)/(3)`, n ∈ Z or θ = nπ, n ∈ Z.