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Question
Find the general solution of the following equation:
4 cos2 θ = 3
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Solution
The general solution of cos2θ = cos2α is θ = nπ ± α, n ∈ Z.
Now, 4 cos2 θ = 3
∴ `cos^2θ = (3)/(4) = (sqrt(3)/2)^2`
∴ `cos^2θ = (cos π/(6))^2` ...`[∵ cos π/6 = sqrt(3)/(2)]`
∴ `cos^2θ = cos^2 π/6`
∴ The required general solution is given by `θ = nπ ± (π)/(6), n ∈ Z`.
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