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Find the general solution of the following equation: 4sin2θ = 1.

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Question

Find the general solution of the following equation:

4sin2θ = 1.

Sum
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Solution

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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Chapter 3: Trigonometric Functions - Exercise 3.1 [Page 75]

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