The solution set of the system of equations x | vedclass

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(B) Given the system of equations:$x + y = \frac{2\pi}{3}$$\cos x + \cos y = \frac{3}{2}$Using the sum-to-product formula $\cos A + \cos B = 2 \cos \f

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null set

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(B) Given the system of equations:$x + y = \frac{2\pi}{3}$$\cos x + \cos y = \frac{3}{2}$Using the sum-to-product formula $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$,we get:$2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} = \frac{3}{2}$Substitute $x+y = \frac{2\pi}{3}$ into the equation:$2 \cos \left(\frac{2\pi/3}{2}\right) \cos \frac{x-y}{2} = \frac{3}{2}$$2 \cos \frac{\pi}{3} \cos \frac{x-y}{2} = \frac{3}{2}$Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have:$2 \left(\frac{1}{2}\right) \cos \frac{x-y}{2} = \frac{3}{2}$$\cos \frac{x-y}{2} = \frac{3}{2}$Since the range of the cosine function is $[-1, 1]$ and $\frac{3}{2} > 1$,there is no real value of $x$ and $y$ that satisfies this equation.Therefore,the solution set is a null set.

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

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