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 Exams — Trigonometry
The value of θ satisfying are
3 cos2 θ – \(2 \sqrt{3}\) sin θ cos θ – 3 sin2 θ = 0 are (n∈I)
3 cos2 θ – \(2 \sqrt{3}\) sin θ cos θ – 3 sin2 θ = 0 are (n∈I)
- \(\mathrm{n} \pi-\frac{2 \pi}{3}, \mathrm{n} \pi+\frac{\pi}{6}\)
- \(\mathrm{n} \pi-\frac{\pi}{3}, \mathrm{n} \pi+\frac{\pi}{6}\)
- \(2 \mathrm{n} \pi-\frac{\pi}{3}, \mathrm{n} \pi\)
- \(2 \mathrm{n} \pi+\frac{\pi}{3}, \mathrm{n} \pi\)
Answer
(B) n - 3 , n + 6
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