If r > 0, –π ≤ θ ≤ π and r, θ satisfy r sin θ = 3 and r = 4 (1 + sin θ), then the number…
Mathematics · JEE Advanced · NTA Exams — Trigonometry
If r > 0, –π ≤ θ ≤ π and r, θ satisfy r sin θ = 3 and
r = 4 (1 + sin θ), then the number of possible solutions of the pair (r, θ) is
r = 4 (1 + sin θ), then the number of possible solutions of the pair (r, θ) is
- 2
- 4
- 0
- inifinite
Answer
(A) 2
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