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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

Which of the following set of values of x satisfies the equation 2^...

Which of the following set of values of xx satisfies the equation 2(2sin⁡2x−3sin⁡x+1)2^{\left(2 \sin ^2 x-3 \sin x+1\right)}+2^{\left(2-2 \sin ^2 x+3 \sin x\right)}=9 ?

Options

  1. Ax=nπ±π6,n∈I\mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{6}, \mathrm{n} \in \mathrm{I}
  2. Bx=nπ±π3,n∈I\mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{3}, \mathrm{n} \in \mathrm{I}
  3. Cx=nπ,n∈I\mathrm{x}=\mathrm{n} \pi, \mathrm{n} \in \mathrm{I}
  4. Dx=2nπ+π2,n∈I\mathrm{x}=2 \mathrm{n} \pi+\frac{\pi}{2}, \mathrm{n} \in \mathrm{I}

Correct answer

D. x=2nπ+π2,n∈I\mathrm{x}=2 \mathrm{n} \pi+\frac{\pi}{2}, \mathrm{n} \in \mathrm{I}

Step-by-step solution

2(2sin⁡2x−3sin⁡x+1)+23−(2sin⁡2x−3sin⁡x+1)=922sin⁡2x−3sin⁡x+1=tt+8t=9⇒t2−9t+8=0⇒(t−8)(t−1)=0⇒2sin⁡2x−3sin⁡x+1=3 or 2sin⁡2x−3sin⁡x+1=0sin⁡x=−12,sin⁡x=12,sin⁡x=1\begin{aligned} & 2^{\left(2 \sin ^2 x-3 \sin x+1\right)}+2^{3-\left(2 \sin ^2 x-3 \sin x+1\right)}=9 \\ & 2^{2 \sin ^2 x-3 \sin x+1}=t \\ & t+\frac{8}{t}=9 \Rightarrow t^2-9 t+8=0 \\ & \Rightarrow \quad(t-8)(t-1)=0 \\ & \Rightarrow \quad 2 \sin ^2 x-3 \sin x+1=3 \\ & \text { or } \quad 2 \sin ^2 x-3 \sin x+1=0 \\ & \sin x=-\frac{1}{2}, \quad \sin x=\frac{1}{2}, \quad \sin x=1\end{aligned}
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