The value of θ satisfying are 3 cos 2 θ – 2 3 sin θ cos θ – 3 sin 2 θ = 0 are (n∈I)

Mathematics · JEE Advanced · NTA ExamsTrigonometry

The value of θ satisfying are
3 cos2 θ – \(2 \sqrt{3}\) sin θ cos θ – 3 sin2 θ = 0 are (n∈I)
  1. \(\mathrm{n} \pi-\frac{2 \pi}{3}, \mathrm{n} \pi+\frac{\pi}{6}\)
  2. \(\mathrm{n} \pi-\frac{\pi}{3}, \mathrm{n} \pi+\frac{\pi}{6}\)
  3. \(2 \mathrm{n} \pi-\frac{\pi}{3}, \mathrm{n} \pi\)
  4. \(2 \mathrm{n} \pi+\frac{\pi}{3}, \mathrm{n} \pi\)

Answer

(B) n - 3 , n + 6

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