Solution of Triangles & Trig EquationsmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Trig Equation Solution Sum = -4π/3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let S={θ∈(−2π,2π):cos⁡θ+1=3sin⁡θ}S=\left\{\theta\in(-2\pi,2\pi):\cos\theta+1=\sqrt3\sin\theta\right\}. Then ∑θ∈Sθ\displaystyle\sum_{\theta\in S}\theta is equal to
A−2π3-\dfrac{2\pi}{3}
B−4π3-\dfrac{4\pi}{3}correct
C2π3\dfrac{2\pi}{3}
D4π3\dfrac{4\pi}{3}
Solution
Step 1: cos⁡θ+1=2cos⁡2θ2\cos\theta+1=2\cos^2\dfrac{\theta}{2}, sin⁡θ=2sin⁡θ2cos⁡θ2\sin\theta=2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}. Then cos⁡θ+1=3sin⁡θ\cos\theta+1=\sqrt3\sin\theta:
2cos⁡2θ2=23sin⁡θ2cos⁡θ2⇒2cos⁡θ2(cos⁡θ2−3sin⁡θ2)=0.2\cos^2\dfrac{\theta}{2}=2\sqrt3\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}\Rightarrow2\cos\dfrac{\theta}{2}\left(\cos\dfrac{\theta}{2}-\sqrt3\sin\dfrac{\theta}{2}\right)=0.
 Step 2: ∙\bullet cos⁡θ2=0⇒θ2=±π2⇒θ=±π\cos\dfrac{\theta}{2}=0\Rightarrow\dfrac{\theta}{2}=\pm\dfrac{\pi}{2}\Rightarrow\theta=\pm\pi. ∙\bullet cos⁡θ2−3sin⁡θ2=0⇒tan⁡θ2=13\cos\dfrac{\theta}{2}-\sqrt3\sin\dfrac{\theta}{2}=0\Rightarrow\tan\dfrac{\theta}{2}=\dfrac{1}{\sqrt3}. Step 3: θ∈(−2π,2π)⇒θ2∈(−π,π)\theta\in(-2\pi,2\pi)\Rightarrow\dfrac{\theta}{2}\in(-\pi,\pi). tan⁡θ2=13\tan\dfrac{\theta}{2}=\dfrac{1}{\sqrt3} at θ2=π6, −5π6\dfrac{\theta}{2}=\dfrac{\pi}{6},\ -\dfrac{5\pi}{6}:
θ=π3, −5π3.\theta=\dfrac{\pi}{3},\ -\dfrac{5\pi}{3}.
 Step 4: S={−5π3, −π, π3, π}S=\left\{-\dfrac{5\pi}{3},\,-\pi,\,\dfrac{\pi}{3},\,\pi\right\}.
∴∑θ∈Sθ=π3−5π3+π−π=π−5π3=−4π3.\therefore\sum_{\theta\in S}\theta=\dfrac{\pi}{3}-\dfrac{5\pi}{3}+\pi-\pi=\dfrac{\pi-5\pi}{3}=-\dfrac{4\pi}{3}.
 Correct answer: (2)
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