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

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प्रश्न

Find the general solution of the following equation:

4sin2θ = 1.

बेरीज
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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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पाठ 3: Trigonometric Functions - Exercise 3.1 [पृष्ठ ७५]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Trigonometric Functions
Exercise 3.1 | Q 6.2 | पृष्ठ ७५

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