Вариант 3 , задание 3, заранее спасибо. Created by qirl4ka. algebra-ru

Вариант 3 , задание 3, заранее спасибо...

qirl4ka @qirl4ka

March 2022 1 16 Report

Вариант 3 , задание 3, заранее спасибо

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dtnth

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$\left \{ {{x+y=\frac{\pi}{2}} \atop {sin x+sin y=-\sqrt{2}}} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {sin x+sin y=-\sqrt{2}}} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {sin x+sin (\frac{\pi}{2}-x)=-\sqrt{2}}} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {sin x+cosx=-\sqrt{2}}} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {\sqrt{2}(\frac{1}{\sqrt{2}}sin x+\frac{1}{\sqrt{2}}cosx)=-\sqrt{2}}} \right;\\$

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