JEE Mains · Maths · STD 11 - Trigonometrical equations
Let \(S = \{\theta \in (-2\pi, 2\pi) : \cos\theta + 1 = \sqrt{3}\sin\theta\}\). Then \(\sum_{\theta \in S}\theta\) is equal to:
- A \(-\dfrac{2\pi}{3}\)
- B \(-\dfrac{4\pi}{3}\)
- C \(\dfrac{2\pi}{3}\)
- D \(\dfrac{4\pi}{3}\)
Answer & Solution
Correct Answer
(B) \(-\dfrac{4\pi}{3}\)
Step-by-step Solution
Detailed explanation
Given equation: \(\cos\theta + 1 = \sqrt{3}\sin\theta\) Using half-angle formulas, we get: \(2\cos^2\dfrac{\theta}{2} = 2\sqrt{3}\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}\) \(2\cos\dfrac{\theta}{2} \left( \cos\dfrac{\theta}{2} - \sqrt{3}\sin\dfrac{\theta}{2} \right) = 0\) This…
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