The solution set of the system of equation
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Question

$\cos x+\cos y=\frac{3}{2}$

$\Rightarrow \quad 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{3}{2}$

$\Rightarrow \quad 2 \cos \frac{\pi}{3} \cos

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Vedclass · Answer

$\cos x+\cos y=\frac{3}{2}$

$\Rightarrow \quad 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{3}{2}$

$\Rightarrow \quad 2 \cos \frac{\pi}{3} \cos \frac{x-y}{2}=\frac{3}{2} \quad\left(\because x+y=\frac{2 \pi}{3}ight)$

$\Rightarrow \quad \cos \frac{x-y}{2}=\frac{3}{2}$

$	ext { which is not possible } \quad\left(\because \frac{3}{2}>1ight)$

Hence the system of equations has no solution.

The solution set of the system of equation

$x\,\, + \,\,y\,\, = \,\,\frac{{2\pi }}{3},\,{\rm{cos}}\,{\rm{x + }}\,{\rm{ cos}}\,{\rm{y}}\,{\rm{ = }}\,\frac{3}{2},$ where $x$ and $y$ are real in

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The solution set of the system of equation $x\,\, + \,\,y\,\, = \,\,\frac{{2\pi }}{3},\,{\rm{cos}}\,{\rm{x + }}\,{\rm{ cos}}\,{\rm{y}}\,{\rm{ = }}\,\frac{3}{2},$ where $x$ and $y$ are real in