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Question

Choose the correct option of general solutions -
Trigonometric EquationsGeneral Solutions1. sin2θ=sin2xA.2nπ±α2. cos2θ=cos2xB.nπ+(1)nα3. tan2θ=tan2xC.nπ±α

A

1 - A, 2 - B, 3 - C

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B

1 - B, 2 - A, 3 - C

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C

1 - A, 2 - A, 3 - A

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D

1 - C, 2 - C, 3 - C

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Solution

The correct option is D

1 - C, 2 - C, 3 - C


1 . sin2θ = sin2α

sin2θ - sin2α = 0

= sin (θ+α) × sin (θ-α) = 0

So, sin (θ+α) = 0 or sin (θ-α) = 0

θ + α = nπ or θ - α = nπ

θ = nπ - α or θ = nπ + α

Combining both the equations,

We get, θ = nπ ± α

2. cos2θ = cos2α

cos2θ - cos2α = 0

- sin (θ+α) sin (θ-α) = 0

sin (θ+α) =0 or sin (θ-α) = 0

θ + α = nπ or θ - α = nπ

θ = nπ - α or θ = nπ + α

Combining both the equations,

We get, θ = nπ ± α

3. tan2θ = tan2α

sin2θcos2θ - sin2αcos2α = 0

sin2θ.cos2αcos2θ.sin2αcos2θ.cos2α = 0

(sinθ.cosα)2(cosθ.sinα)2cos2θ.cos2α = 0

(sinθ.cosα+cosθ.sinα)(sinθ.cosαcosθ.sinα)cos2θ.cos2α = 0

sin(θ +α) sin(θ -α) = 0 { where θ&α should not be equal to (2n+1)π2 n ∈ Z}

sin (θ+α) =0 or sin (θ-α) = 0

θ + α = nπ or θ - α = nπ

θ = nπ - α or θ = nπ + α

Combining both the equations,

We get, θ = nπ ± α

Where n ∈ Z, α & θ should not be equal odd multiples of π /2

α & θ ≠ (2n+1) π /2 ∈ Z


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