Question

Помогите решить систему cosx*siny=корень2/2 и x+y=3П/4

Answer 1

Ответ:

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

Объяснение:

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

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

Тогда получим два случая:

1)

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

2)

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