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JEE Main 201815 AprilEvening ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Evening Shift
Mode
Online
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2018 · 15 April · Evening Shift · Online · Mathematics · Trigonometric Equations

The number of solutions of 3 x= 2 x , in the interval ((π)/(2), π)...

The number of solutions of sin⁡3x=cos⁡2x\sin 3 x=\cos 2 x, in the interval (π2,π)\left(\frac{\pi}{2}, \pi\right) is

Options

  1. A3
  2. B4
  3. C2
  4. D1

Correct answer

D. 1

Step-by-step solution

\sin 3 x=\cos 2 x ⇒3sin⁡x−4sin⁡3x=1−2sin⁡2x⇒4sin⁡3x−2sin⁡2x−3sin⁡x+1=0⇒sin⁡x=1,−2±258\begin{aligned}&\Rightarrow 3 \sin x-4 \sin ^3 x=1-2 \sin ^2 x \\ &\Rightarrow 4 \sin ^3 x-2 \sin ^2 x-3 \sin x+1=0 \\ &\Rightarrow \sin x=1, \frac{-2 \pm 2 \sqrt{5}}{8} \end{aligned}Intheinterval(π2,π),sin⁡x=−2+258In the interval \left(\frac{\pi}{2}, \pi\right), \sin x=\frac{-2+2 \sqrt{5}}{8} So, there is only one solution.
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