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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Trigonometric Equations

The sum of all values of θ ∈[0,2 π] satisfying 2 ^2 θ= 2 θ and 2 ^2...

The sum of all values of θ∈[0,2π]\theta \in[0,2 \pi] satisfying 2sin⁡2θ=cos⁡2θ2 \sin ^2 \theta=\cos 2 \theta and 2cos⁡2θ=3sin⁡θ2 \cos ^2 \theta=3 \sin \theta is

Options

  1. A4π4 \pi
  2. B5π6\frac{5 \pi}{6}
  3. Cπ\pi
  4. Dπ2\frac{\pi}{2}

Correct answer

C. π\pi

Step-by-step solution

2sin⁡2θ=cos⁡2θ2sin⁡2θ=1−2sin⁡2θ4sin⁡2θ=1sin⁡2θ=14sin⁡θ=±122cos⁡2θ=3sin⁡θ2−2sin⁡2θ+3sin⁡θ−2=0(2sin⁡θ−1)(2sin⁡θ−2)=0sin⁡θ=12\begin{aligned} & 2 \sin ^2 \theta=\cos 2 \theta \\ & 2 \sin ^2 \theta=1-2 \sin ^2 \theta \\ & 4 \sin ^2 \theta=1 \\ & \sin ^2 \theta=\frac{1}{4} \\ & \sin \theta= \pm \frac{1}{2} \\ & 2 \cos ^2 \theta=3 \sin \theta \\ & 2-2 \sin ^2 \theta+3 \sin \theta-2=0 \\ & (2 \sin \theta-1)(2 \sin \theta-2)=0 \\ & \sin \theta=\frac{1}{2} \end{aligned} so common equation which satisfy both equations is sin⁡θ=12θ=π6,5π6(θ∈[0,2π])\sin \theta=\frac{1}{2} \theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad(\theta \in[0,2 \pi]) Sum =π=\pi
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