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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

If 6 θ=32 ^5 θ θ-32 ^3 θ θ+3 x then x is equal to :

If sin⁡6θ=32cos⁡5θsin⁡θ−32cos⁡3θsin⁡θ+3x\sin 6 \theta=32 \cos ^5 \theta \sin \theta-32 \cos ^3 \theta \sin \theta+3 x then xx is equal to :

Options

  1. Acos⁡θ\cos \theta
  2. Bcos⁡2θ\cos 2 \theta
  3. Csin⁡θ\sin \theta
  4. Dsin⁡2θ\sin 2 \theta

Correct answer

D. sin⁡2θ\sin 2 \theta

Step-by-step solution

We have, sin⁡6θ=sin⁡3(2θ)=3sin⁡2θ−4sin⁡32θ=3⋅2sin⁡θcos⁡θ−4⋅8⋅cos⁡3θsin⁡3θ=6sin⁡θcos⁡θ−32cos⁡3θsin⁡θ(1−cos⁡2θ)sin⁡6θ=32cos⁡5θ⋅sin⁡θ−32cos⁡3θsin⁡θ+3sin⁡2θ…\sin 6 \theta=\sin 3(2 \theta)=3 \sin 2 \theta-4 \sin ^3 2 \theta=3 \cdot 2 \sin \theta \cos \theta-4 \cdot 8 \cdot \cos ^3 \theta \sin ^3 \theta=6 \sin \theta \cos \theta-32 \cos ^3 \theta \sin \theta\left(1-\cos ^2 \theta\right)\sin 6 \theta=32 \cos ^5 \theta \cdot \sin \theta-32 \cos ^3 \theta \sin \theta+3 \sin 2 \theta \ldots (i) Given that, sin⁡6θ=32cos⁡5θsin⁡θ−32cos⁡3θsin⁡θ+3x…\sin 6 \theta=32 \cos ^5 \theta \sin \theta-32 \cos ^3 \theta \sin \theta+3 x\ldots (ii) On comparing Eqs. (i) and (ii), we get 3x=3sin⁡2θ⇒x=sin⁡2θ3 x=3 \sin 2 \theta \Rightarrow x=\sin 2 \theta
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