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MathMedMain 2026 • 6th Apr - Shift 1

Sum of Solutions for Trigonometric Equation in Interval

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

Options

A

−2π3-\frac{2\pi}{3}

B

−4π3-\frac{4\pi}{3}

Correct
C

2π3\frac{2\pi}{3}

D

4π3\frac{4\pi}{3}

Topics & Concepts

Trigonometric FunctionsTrigonometric Equations

Step-by-Step Solution

To find the sum of all values of θ∈(−2π,2π)\theta \in (-2\pi, 2\pi) satisfying the given trigonometric equation, we begin by writing the equation:

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

Rearranging the terms, we get: 3sin⁡θ−cos⁡θ=1\sqrt{3} \sin \theta - \cos \theta = 1

Dividing both sides of the equation by 22: 32sin⁡θ−12cos⁡θ=12\frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta = \frac{1}{2}

Using the trigonometric identity for sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A - B) = \sin A \cos B - \cos A \sin B, where cos⁡(π6)=32\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} and sin⁡(π6)=12\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, we can rewrite the left-hand side as: sin⁡(θ−π6)=12\sin \left(\theta - \frac{\pi}{6}\right) = \frac{1}{2}

The general solution for this equation is: θ−π6=nπ+(−1)nπ6,n∈Z\theta - \frac{\pi}{6} = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z}

Now, we test integer values for nn to find solutions for θ\theta in the interval (−2π,2π)(-2\pi, 2\pi):

  1. For n=0n = 0: θ−π6=π6  ⟹  θ=π3∈(−2π,2π)\theta - \frac{\pi}{6} = \frac{\pi}{6} \implies \theta = \frac{\pi}{3} \in (-2\pi, 2\pi)

  2. For n=1n = 1: θ−π6=π−π6=5π6  ⟹  θ=π∈(−2π,2π)\theta - \frac{\pi}{6} = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \implies \theta = \pi \in (-2\pi, 2\pi)

  3. For n=2n = 2: θ−π6=2π+π6  ⟹  θ=7π3∉(−2π,2π)\theta - \frac{\pi}{6} = 2\pi + \frac{\pi}{6} \implies \theta = \frac{7\pi}{3} \notin (-2\pi, 2\pi)

  4. For n=−1n = -1: θ−π6=−π−π6=−7π6  ⟹  θ=−π∈(−2π,2π)\theta - \frac{\pi}{6} = -\pi - \frac{\pi}{6} = -\frac{7\pi}{6} \implies \theta = -\pi \in (-2\pi, 2\pi)

  5. For n=−2n = -2: θ−π6=−2π+π6=−11π6  ⟹  θ=−5π3∈(−2π,2π)\theta - \frac{\pi}{6} = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6} \implies \theta = -\frac{5\pi}{3} \in (-2\pi, 2\pi)

  6. For n=−3n = -3: θ−π6=−3π−π6  ⟹  θ=−3π∉(−2π,2π)\theta - \frac{\pi}{6} = -3\pi - \frac{\pi}{6} \implies \theta = -3\pi \notin (-2\pi, 2\pi)

Thus, the set of solutions SS in the given interval (−2π,2π)(-2\pi, 2\pi) is: S={−5π3,−π,π3,π}S = \left\{ -\frac{5\pi}{3}, -\pi, \frac{\pi}{3}, \pi \right\}

The sum of all elements in SS is: ∑θ∈Sθ=(−5π3)+(−π)+π3+π=−4π3\sum_{\theta \in S} \theta = \left(-\frac{5\pi}{3}\right) + (-\pi) + \frac{\pi}{3} + \pi = -\frac{4\pi}{3}

Hence, the correct option is B.

Sum of Solutions for Trigonometric Equation in Interval | Mathematics PYQ Solution - JEE Challenger