Q39 / 75 · JEE Main 2026 (06 April Shift 1) · Mathematics

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
    $-\dfrac{2\pi}{3}$
  • Correct answer
    B
    $-\dfrac{4\pi}{3}$
  • C
    $\dfrac{2\pi}{3}$
  • D
    $\dfrac{4\pi}{3}$

Solution

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 gives two cases:

Case 1: $\cos\dfrac{\theta}{2} = 0$

$\Rightarrow \dfrac{\theta}{2} = (2k+1)\dfrac{\pi}{2} \Rightarrow \theta = (2k+1)\pi$

Since $\theta \in (-2\pi, 2\pi)$, the possible values are $\theta = -\pi, \pi$.

Case 2: $\cos\dfrac{\theta}{2} - \sqrt{3}\sin\dfrac{\theta}{2} = 0$

$\Rightarrow \tan\dfrac{\theta}{2} = \dfrac{1}{\sqrt{3}}$

$\Rightarrow \dfrac{\theta}{2} = k\pi + \dfrac{\pi}{6} \Rightarrow \theta = 2k\pi + \dfrac{\pi}{3}$

Since $\theta \in (-2\pi, 2\pi)$, the possible values are:

For $k = 0$, $\theta = \dfrac{\pi}{3}$

For $k = -1$, $\theta = -2\pi + \dfrac{\pi}{3} = -\dfrac{5\pi}{3}$

Thus, the set of solutions is $S = \left\{-\pi, \pi, \dfrac{\pi}{3}, -\dfrac{5\pi}{3}\right\}$.

Sum of all solutions in $S$:

$\sum_{\theta \in S}\theta = -\pi + \pi + \dfrac{\pi}{3} - \dfrac{5\pi}{3} = -\dfrac{4\pi}{3}$

Answer: $-\dfrac{4\pi}{3}$

Hence, the correct answer is option: 'B'.

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