The set of solutions of the system of equatio | vedclass

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(D) Given the system of equations: $x+y = \frac{2 \pi}{3}$ $(i)$ $\cos x + \cos y = \frac{3}{2}$ (ii) Using the sum-to-product formula,$\cos x + \cos

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Empty set

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(D) Given the system of equations: $x+y = \frac{2 \pi}{3}$ $(i)$ $\cos x + \cos y = \frac{3}{2}$ (ii) Using the sum-to-product formula,$\cos x + \cos y = 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$. Substituting $(i)$ into the formula: $2 \cos \left(\frac{1}{2} \cdot \frac{2 \pi}{3}\right) \cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$ $2 \cos \left(\frac{\pi}{3}\right) \cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$ Since $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$,we have: $2 \left(\frac{1}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$ $\cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$ Since the range of the cosine function is $[-1, 1]$,the equation $\cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$ has no real solution. Therefore,the system of equations has an empty set of solutions.

The set of solutions of the system of equations $x+y = \frac{2 \pi}{3}$ and $\cos x + \cos y = \frac{3}{2}$,where $x, y$ are real,is

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