решить систему уравнений: а) {cos x + cos y = 1; {x + y = 2П б) {sin x = cos y; {2 cos...

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$\\ a)\quad\left\{\begin{matrix} \cos x+\cos y=1\\ x+y=2\pi \end{matrix}\right.\Rightarrow \left\{\begin{matrix} \cos\left ( 2\pi-y \right )+\cos y=1\\ x=2\pi-y \end{matrix}\right.\Rightarrow \left\{\begin{matrix} \cos y+\cos y=1\\ x=2\pi-y \end{matrix}\right.\Rightarrow\\ \Rightarrow \left\{\begin{matrix} 2\cos y=1\\ x=2\pi-y \end{matrix}\right.\Rightarrow \left\{\begin{matrix} \cos y=\frac12\\ x=2\pi-y \end{matrix}\right.\Rightarrow \left\{\begin{matrix} y=\frac{\pi}3+2\pi n\\ x=\frac{5\pi}{3}+2\pi n \end{matrix}\right.\\b)\quad\left\{\begin{matrix}\sin x =\cos y\\2\cos^2y+\sin x=3\end{matrix}\right.\Rightarrow \left\{\begin{matrix}\sin x=\cos y\\2\cos^2y+\cos y-3=0\end{matrix}\right.\\2\cos^2y+\cos y-3=0\\\cos y=t,t\in\left [ -1,1 \right ]\\2t^2+t-3=0\\D=1-4\cdot2\cdot(-3)=25\\t_1=1,\quad t_2=-\frac32\\\cos y=1\Rightarrow y=2\pi n\\\left\{\begin{matrix}\sin x=1\\ y=2\pi n\end{matrix}\right.\Rightarrow \left\{\begin{matrix}x=\frac{\pi}{2}+2\pi n\\ y=2\pi n\end{matrix}\right.$