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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

If 2 ^2 θ+2 2 =3 cosec^2 θ , where θ ∈(0, π) , then:

If 2sin⁡2θ+22=3cosec⁡2θ2 \sin ^2 \theta+2 \sqrt{2}=3 \operatorname{cosec}^2 \theta, where θ∈(0,π)\theta \in(0, \pi), then:

Options

  1. Anumber of real solutions is 2 .
  2. Bsum of all solutions is π\pi.
  3. Cnumber of real solution is 4 .
  4. Dsum of all solutions is 4π4 \pi.

Correct answer

C. number of real solution is 4 .

Step-by-step solution

2t2+22t=3;t=sin⁡2θ∴t=sin⁡2θ=12⇒sin⁡θ=±1242 t^2+2 \sqrt{2} t=3 ; t=\sin ^2 \theta\therefore \quad t=\sin ^2 \theta=\frac{1}{\sqrt{2}} \quad \Rightarrow \quad \sin \theta= \pm \frac{1}{\sqrt[4]{2}}
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