JEE Main 2026 — Mathematics Trigonometry

Given equation: $\cos\theta + 1 = \sqrt{3}\sin\theta$cosθ+1=3​sinθ

Using half-angle formulas, we get:

$$2\cos^2\dfrac{\theta}{2} = 2\sqrt{3}\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}$$2cos22θ​=23​sin2θ​cos2θ​

$$2\cos\dfrac{\theta}{2} \left( \cos\dfrac{\theta}{2} - \sqrt{3}\sin\dfrac{\theta}{2} \right) = 0$$2cos2θ​(cos2θ​−3​sin2θ​)=0

This gives two cases:

Case 1: $\cos\dfrac{\theta}{2} = 0$cos2θ​=0

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

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

Case 2: $\cos\dfrac{\theta}{2} - \sqrt{3}\sin\dfrac{\theta}{2} = 0$cos2θ​−3​sin2θ​=0

$$\Rightarrow \tan\dfrac{\theta}{2} = \dfrac{1}{\sqrt{3}}$$⇒tan2θ​=3​1​

$$\Rightarrow \dfrac{\theta}{2} = k\pi + \dfrac{\pi}{6} \Rightarrow \theta = 2k\pi + \dfrac{\pi}{3}$$⇒2θ​=kπ+6π​⇒θ=2kπ+3π​

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

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

For $k = -1$k=−1, $\theta = -2\pi + \dfrac{\pi}{3} = -\dfrac{5\pi}{3}$θ=−2π+3π​=−35π​

Thus, the set of solutions is $S = \left\{-\pi, \pi, \dfrac{\pi}{3}, -\dfrac{5\pi}{3}\right\}$S={−π,π,3π​,−35π​}.

Sum of all solutions in $S$S:

$$\sum_{\theta \in S}\theta = -\pi + \pi + \dfrac{\pi}{3} - \dfrac{5\pi}{3} = -\dfrac{4\pi}{3}$$θ∈S∑​θ=−π+π+3π​−35π​=−34π​

Answer: $-\dfrac{4\pi}{3}$−34π​