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Mathematics · Trigonometry Ratios and Identities

JEE Main 2025 — 2 April, Morning Shift — Question 38

If θ∈[−2π,2π]\theta \in[-2 \pi, 2 \pi], then the number of solutions of 22cos⁡2θ+(2−6)cos⁡θ−3=02 \sqrt{2} \cos ^{2} \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0, is equal to

  1. Option A:

    1010

  2. Option B:

    66

  3. Option C:

    88

    Correct
  4. Option D:

    1212

Answer: C

Step-by-step solution

22cos⁡2θ+(2−6)cos⁡θ−3=02 \sqrt{2} \cos ^{2} \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0

22cos⁡2θ+2cos⁡θ−6cos⁡θ−3=0(2cos⁡θ−3)(2cos⁡θ+1)=0⇒cos⁡θ=32 or cos⁡θ=−12θ={−11π6,−5π4,−3π4,−π6,π6,3π4,5π4,11π6}⇒8 (solution) \begin{aligned} & 2 \sqrt{2} \cos ^{2} \theta+2 \cos \theta-\sqrt{6} \cos \theta-\sqrt{3}=0 \\& (2 \cos \theta-\sqrt{3})(\sqrt{2} \cos \theta+1)=0 \\& \Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \text { or } \cos \theta=\frac{-1}{\sqrt{2}} \\& \theta=\left\{\frac{-11 \pi}{6}, \frac{-5 \pi}{4}, \frac{-3 \pi}{4}, \frac{-\pi}{6}, \frac{\pi}{6}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{11 \pi}{6}\right\} \\& \Rightarrow 8 \text { (solution) } \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
2 April, Morning Shift
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Periodicity of trigometric functions,Solutions of trigonometric equations