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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Trigonometric Equations

The number of solutions of equation (4- 3 ) x -2 3 ^2 x=- 4 1+ 3 ,...

The number of solutions of equation (4−3)sin⁡x(4-\sqrt{3}) \sin x −23cos⁡2x=−41+3,x∈[−2π,5π2]-2 \sqrt{3} \cos ^2 x=-\frac{4}{1+\sqrt{3}}, x \in\left[-2 \pi, \frac{5 \pi}{2}\right] is

Options

  1. A44
  2. B33
  3. C66
  4. D55

Correct answer

D. 55

Step-by-step solution

(4−3)sin⁡x−23cos⁡2x=−41+3,x∈[−2π,5π2]⇒(4−3)sin⁡x−23(1−sin⁡2x)=2(1−3)⇒23sin⁡2x+4sin⁡x−3sin⁡x−2=0⇒(2sin⁡x−1)(3sin⁡x+2)=0⇒sin⁡x=12∴ Number of  solution =5\begin{aligned} & (4-\sqrt{3}) \sin x-2 \sqrt{3} \cos ^2 x=\frac{-4}{1+\sqrt{3}}, x \in\left[-2 \pi, \frac{5 \pi}{2}\right] \\ & \Rightarrow(4-\sqrt{3}) \sin x-2 \sqrt{3}\left(1-\sin ^2 x\right)=2(1-\sqrt{3}) \\ & \Rightarrow 2 \sqrt{3} \sin ^2 x+4 \sin x-\sqrt{3} \sin x-2=0 \\ & \Rightarrow(2 \sin x-1)(\sqrt{3} \sin x+2)=0 \\ & \Rightarrow \sin x=\frac{1}{2} \\ & \therefore \text{ Number\text{ of } solution }=5\end{aligned}
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