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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

If x+1=(1)/(2+(1)/(2+(1)/(2+ ∞))) where x ∈(0,2 π) , then:

If sin⁡x+1=12+12+12+……∞\sin x+1=\frac{1}{2+\frac{1}{2+\frac{1}{2+\ldots \ldots \infty}}} where x∈(0,2π)x \in(0,2 \pi), then:

Options

  1. Asum of all the solutions of the equation is 3π3 \pi.
  2. Bsum of all the solutions of the equation is 6π6 \pi.
  3. Cnumber of solutions of the equation is 2 .
  4. Dnumber of solutions of the equation is 4 .

Correct answer

C. number of solutions of the equation is 2 .

Step-by-step solution

sin⁡x+1=12+sin⁡x+1\sin x+1=\frac{1}{2+\sin x+1}⇒(sin⁡x+1)(sin⁡x+3)=1⇒sin⁡2x+4sin⁡x+2=0⇒(sin⁡x+2)2=2⇒sin⁡x=−2±2⇒sin⁡x=−2−2 (rejected) ⇒sin⁡x=−2+2 (negative) ⇒π+α,2π−α\begin{aligned} & \Rightarrow \quad(\sin x+1)(\sin x+3)=1 \quad \Rightarrow \quad \sin ^2 x+4 \sin x+2=0 \\ & \Rightarrow \quad(\sin x+2)^2=2 \quad \Rightarrow \quad \sin x=-2 \pm \sqrt{2} \\ & \Rightarrow \sin x=-2-\sqrt{2} \quad \text { (rejected) } \\ & \Rightarrow \sin x=-2+\sqrt{2} \text { (negative) } \quad \Rightarrow \quad \pi+\alpha, 2 \pi-\alpha\end{aligned}$$
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