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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

The set of solutions of the system of equations aligned x+y & =(2...

The set of solutions of the system of equations x+y=2π3 and \cosx+cos⁡y=32,\begin{aligned}x+y & =\frac{2 \pi}{3}\\ \text { and } \quad \cosx+\cos y & =\frac{3}{2}, \end{aligned}where x,yx, y are real, is

Options

  1. A\left\{(x, y): \cos \left(\frac{x-y}{2}\right)=\frac{1}{2}\right\}
  2. B\left\{(x, y): \sin \left(\frac{x-y}{2}\right)=\frac{1}{2}\right\}
  3. C\left\{(x, y): \cos (x-y)=\frac{1}{2}\right\}
  4. DEmpty set

Correct answer

D. Empty set

Step-by-step solution

Given system of equation is x+y=2π3x+y=\frac{2 \pi}{3}and cos⁡x+cos⁡y=32\cos x+\cos y=\frac{3}{2}, where x,yx, y are real. ⇒2cos⁡(x+y2)⋅cos⁡(x−y2)=32⇒cos⁡(12⋅2π3)⋅cos⁡(x−y2)=34\begin{aligned}& \Rightarrow \quad 2 \cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)=\frac{3}{2}\\ & \Rightarrow \quad \cos \left(\frac{1}{2} \cdot \frac{2 \pi}{3}\right) \cdot \cos \left(\frac{x-y}{2}\right)=\frac{3}{4} \end{aligned}[fromEq.(i)]from Eq. (i)]⇒cos⁡(π3)⋅cos⁡(x−y2)=34⇒12cos⁡(x−y2)=34⇒cos⁡(x−y2)=32\begin{aligned} \Rightarrow& \cos \left(\frac{\pi}{3}\right) \cdot \cos \left(\frac{x-y}{2}\right)& =\frac{3}{4}\\ \Rightarrow & \frac{1}{2} \cos \left(\frac{x-y}{2}\right)& =\frac{3}{4}\\ \Rightarrow & \cos \left(\frac{x-y}{2}\right)& =\frac{3}{2} \end{aligned}Now,wehaveNow, we havecos⁡(x−y)=2cos⁡2(x−y2)−1=2×\ffrac94−1=\ffrac92−1=72 [from Eq. (ii) ] and cos⁡(x−y)=1−2sin⁡2(x−y2)⇒sin⁡(x−y2)<0\begin{aligned}& \cos (x-y)=2 \cos ^2\left(\frac{x-y}{2}\right)-1 \\ & \left.=2 \times \ffrac{9}{4}-1=\ffrac{9}{2}-1=\frac{7}{2} \quad \text { [from Eq. (ii) }\right]\\ & \text { and } \cos (x-y)=1-2 \sin ^2\left(\frac{x-y}{2}\right)\\ & \Rightarrow \quad \sin \left(\frac{x-y}{2}\right)< 0 \\ & \end{aligned}$So, system of equation have empty set of solution.
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