Question

Решить систему cosx*siny=корень2/2 и x+y=3п/4

Answer 1

$(\frac{\pi }{4} +\pi n; \frac{\pi }{2} +\pi n), ( -\pi k; \frac{3\pi }{4} +\pi k) , ~n,k\in\mathbb {Z}$

Объяснение:

$\left \{ \begin{array}{lcl} {{siny*cosx=\frac{\sqrt{2} }{2,} } \\ {x+y=\frac{3\pi }{4}; }} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{\frac{1}{2} (sin(y+x)+sin(y-x))=\frac{\sqrt{2} }{2}, } \\ {x+y=\frac{3\pi }{4} ;}} \end{array} \right.\Leftrightarrow \\ \left \{ \begin{array}{lcl} {{sin\frac{3\pi }{4} +sin (y-x)=\sqrt{2}, } \\ {x+y=\frac{3\pi }{4}; }} \end{array} \right. \Leftrightarrow$

$\left \{ \begin{array}{lcl} {{\frac{\sqrt{2} }{2}+sin (y-x)=\sqrt{2} , } \\ {x+y=\frac{3\pi }{4} ;}} \end{array} \right.\Leftrightarrow \left \{ \begin{array}{lcl} {{sin(y-x)=\frac{\sqrt{2} }{2} ,} \\ {x+y=\frac{3\pi }{4}. }} \end{array} \right.$

Тогда получим два случая:

1)

$\left \{ \begin{array}{lcl} {{y-x=\frac{\pi }{4} +2\pi n,~n\in\mathbb {Z} } \\ {y+x=\frac{3\pi }{4} };} \end{array} \right. \Leftrightarrow \left \{ \begin{array}{lcl} {{2y=\pi +2\pin,~n\in\mathbb {Z} } \\ {2x=\frac{\pi }{2} +2\pi n,~n\in\mathbb {Z}}} \end{array} \right. \Leftrightarrow \\\left \{ \begin{array}{lcl} {{y=\frac{\pi }{2}+\pi n,~n\in\mathbb {Z} } \\ {x=\frac{\pi }{4} + \pi n,~n\in\mathbb {Z}. }} \end{array} \right.$

2)

$\left \{ \begin{array}{lcl} {{y-x=\frac{3\pi }{4}+2\pi k, ~k\in\mathbb {Z} } \\ {y+x=\frac{3\pi }{4}; }} \end{array} \right.\Leftrightarrow \left \{ \begin{array}{lcl} {{2y=\frac{3\pi }{2}+2\pi k,~k\in\mathbb {Z} } \\ {2x=-2\pi k,~k\in\mathbb {Z} ;}} \end{array} \right. \Leftrightarrow\\\left \{ \begin{array}{lcl} {{y=\frac{3\pi }{4}+\pi k, ~k\in\mathbb {Z}} \\ {x=-\pi k ,~k\in\mathbb {Z} }} \end{array} \right.$