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