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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

If a x+ x+1=cosec x then the possible values of x can be

If acos⁡x+cot⁡x+1=cosec⁡xa \cos x+\cot x+1=\operatorname{cosec} x then the possible values of xx can be

Options

  1. Ax=nπ,a∈R,n∈Ix=n \pi, a \in \mathbb{R}, n \in I
  2. Bx=(4n+1)π2,a∈R,n∈Ix=(4 n+1) \frac{\pi}{2}, a \in \mathbb{R}, n \in I
  3. Cx=12sin⁡−1(4(a+1)a2),a∈(−∞,2−22]∪[2+22,∞)−{−1}x=\frac{1}{2} \sin ^{-1}\left(\frac{4(a+1)}{a^2}\right), a \in(-\infty, 2-2 \sqrt{2}] \cup[2+2 \sqrt{2}, \infty)-\{-1\}
  4. Dnone of these

Correct answer

C. x=12sin⁡−1(4(a+1)a2),a∈(−∞,2−22]∪[2+22,∞)−{−1}x=\frac{1}{2} \sin ^{-1}\left(\frac{4(a+1)}{a^2}\right), a \in(-\infty, 2-2 \sqrt{2}] \cup[2+2 \sqrt{2}, \infty)-\{-1\}

Step-by-step solution

sin⁡x+cos⁡x=1−asin⁡xcos⁡x⇒a2sin⁡2xcos⁡2x−2(a+1)sin⁡xcos⁡x=0⇒sin⁡2x(a22sin⁡2x−2(a+1))=0\begin{aligned} & \sin x+\cos x=1-a \sin x \cos x \\ & \Rightarrow a^2 \sin ^2 x \cos ^2 x-2(a+1) \sin x \cos x=0 \\ & \Rightarrow \sin 2 x\left(\frac{a^2}{2} \sin 2 x-2(a+1)\right)=0\end{aligned} Hence sin⁡2x=0∀a∈R⇒x=nπ2,n∈I\sin 2 x=0 \forall a \in R\Rightarrow x=\frac{n \pi}{2}, n \in I And $\sin 2 x=\frac{4(a+1)}{a^2} \Rightarrow a \in(-\infty, 2-2 \sqrt{2}] \cup[2+2 \sqrt{2}, \infty)
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