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JEE Main 20266 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2026 · 6 April · Morning Shift · Mathematics · Trigonometric Equations

Let S = \ θ ∈ (-2π, 2π) : θ + 1 = 3 θ\ . Then ∑_ θ ∈ S θ is equal...

Let S={θ∈(−2π,2π):cos⁡θ+1=3sin⁡θ}S = \{\theta \in (-2\pi, 2\pi) : \cos\theta + 1 = \sqrt{3}\sin\theta\}. Then ∑θ∈Sθ\sum_{\theta \in S}\theta is equal to:

Options

  1. A−2π3-\dfrac{2\pi}{3}
  2. B−4π3-\dfrac{4\pi}{3}
  3. C2π3\dfrac{2\pi}{3}
  4. D4π3\dfrac{4\pi}{3}

Correct answer

B. −4π3-\dfrac{4\pi}{3}

Step-by-step solution

Given equation: cos⁡θ+1=3sin⁡θ\cos\theta + 1 = \sqrt{3}\sin\theta

Using half-angle formulas, we get:

2cos⁡2θ2=23sin⁡θ2cos⁡θ22\cos^2\dfrac{\theta}{2} = 2\sqrt{3}\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}

2cos⁡θ2(cos⁡θ2−3sin⁡θ2)=02\cos\dfrac{\theta}{2} \left( \cos\dfrac{\theta}{2} - \sqrt{3}\sin\dfrac{\theta}{2} \right) = 0

This gives two cases:

Case 1: cos⁡θ2=0\cos\dfrac{\theta}{2} = 0

⇒θ2=(2k+1)π2⇒θ=(2k+1)π\Rightarrow \dfrac{\theta}{2} = (2k+1)\dfrac{\pi}{2} \Rightarrow \theta = (2k+1)\pi

Since θ∈(−2π,2π)\theta \in (-2\pi, 2\pi), the possible values are θ=−π,π\theta = -\pi, \pi.

Case 2: cos⁡θ2−3sin⁡θ2=0\cos\dfrac{\theta}{2} - \sqrt{3}\sin\dfrac{\theta}{2} = 0

⇒tan⁡θ2=13\Rightarrow \tan\dfrac{\theta}{2} = \dfrac{1}{\sqrt{3}}

⇒θ2=kπ+π6⇒θ=2kπ+π3\Rightarrow \dfrac{\theta}{2} = k\pi + \dfrac{\pi}{6} \Rightarrow \theta = 2k\pi + \dfrac{\pi}{3}

Since θ∈(−2π,2π)\theta \in (-2\pi, 2\pi), the possible values are:

For k=0k = 0, θ=π3\theta = \dfrac{\pi}{3}

For k=−1k = -1, θ=−2π+π3=−5π3\theta = -2\pi + \dfrac{\pi}{3} = -\dfrac{5\pi}{3}

Thus, the set of solutions is S={−π,π,π3,−5π3}S = \left\{-\pi, \pi, \dfrac{\pi}{3}, -\dfrac{5\pi}{3}\right\}.

Sum of all solutions in SS:

∑θ∈Sθ=−π+π+π3−5π3=−4π3\sum_{\theta \in S}\theta = -\pi + \pi + \dfrac{\pi}{3} - \dfrac{5\pi}{3} = -\dfrac{4\pi}{3}

Answer: −4π3-\dfrac{4\pi}{3}
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