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प्रश्न
Find the general solution of the following equation:
4sin2θ = 1.
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उत्तर
The general solution of sin2θ = sin2α is
θ = nπ ± α, n ∈ Z.
Now, 4 sin2θ = 1
∴ sin2θ = `1/4 = (1/2)^2`
∴ sin2θ = `(sin pi/6)^2 ...[∵ sin pi/(6) = (1)/(2)]`
∴ sin2θ = sin2 `pi/(6)`
∴ the required general solution is θ = nπ ± `pi/(6)`, n ∈ Z.
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