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प्रश्न
Find the principal solutions of the following equation:
tan 3θ = - 1
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उत्तर
tan 3θ = - 1
Since, θ ∈ (0, 2π), 3θ ∈ (0, 6π)
`tan 3θ = - 1 = - tan π/4 = tan (π - π/4)`
`= tan (2π - π/4) = tan(3π - π/4)`
`= tan(4π - π/4) = tan(5π - π/4)`
`= tan(6π - π/4)` ......[∵ tan (π - θ) = tan(2π - θ) = tan(3π - θ) = tan(4π - θ) = tan (5π - θ) = tan (6π - θ) = - tan θ]
∴ `"tan" 3θ = "tan" (3π)/4 = "tan" (7π)/4 = "tan" (11π)/4 = "tan" (15π)/4 = "tan" (19π)/4 = "tan" (23π)/4`
∴ `3θ = (3π)/4 or 3θ = (7π)/4 or 3θ = (11π)/4 or 3θ = (15π)/4 or 3θ = (19π)/4 or 3θ = (23π)/4`
∴ `θ = π/4 or θ = (7π)/12 or θ = (11π)/12 or θ = (5π)/4 or θ = (19π)/12 or θ = (23π)/12`
Hence, the required principal solutions are,
`{π/4, (7π)/12, (11π)/12, (5π)/4, (19π)/12, (23π)/12}`.
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