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Question
Find the general solutions of the following equation:
`tan^2 theta = 3`
Solution
The general solution of tan2θ = tan2α is θ = nπ ± α, n ∈ Z.
Now, tan2 θ = 3 = `(sqrt3)^2`
∴ tan2 θ = `(tan pi/3)^2 ....[because "tan" pi/3 = sqrt3]`
∴ tan2 θ = `tan^2 (pi)/3`
∴ the required general solution is
∴ θ = `"n"pi +- (pi)/3,` n ∈ Z.
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