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Question
Find the general solution of the following equation:
tan3θ = 3 tanθ.
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Solution
tan3θ = 3tanθ
∴ tan3θ - 3tanθ = 0
∴ tanθ (tan2θ - 3) = 0
∴ either tanθ = 0 or tan2θ - 3 = 0
∴ either tanθ = 0 or tan2θ = 3
∴ either tanθ = 0 or tan2θ = (√3)2
∴ either tan θ = 0 or tan2θ = `(tan pi/3)^2 ...[∵ tan pi/(3) = sqrt(3)]`
∴ either tanθ = 0 or tan2θ = `tan^2 pi/(3)`
The general solution of
tanθ = 0 is θ = nπ, n ∈ Z and
tan2θ = tan2α is θ = nπ ± α, n ∈ Z.
∴ the required general solution is given by
θ = nπ, n ∈ Z or θ = `nπ ± pi/(3)`, n ∈ Z.
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