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प्रश्न
Find the general solution of the following equation:
cos 4θ = cos 2θ
उत्तर
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.
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