Verified answer

Ответ:

[tex]x=\frac{1}{6}(12\pi n-\pi ), x= \frac{1}{6}(12\pi n+\pi ), x=\frac{1}{6}(12\pi n-5\pi ), x=\frac{1}{6}(12\pi n+5\pi )[/tex]

[tex]y=\frac{1}{6}(12\pi n-\pi ), y= \frac{1}{6}(12\pi n+\pi ), y=\frac{1}{6}(12\pi n-5\pi ),y=\frac{1}{6}(12\pi n+5\pi )[/tex],n∈N

Пошаговое объяснение:

[tex]\left \{ {cos(x-y)=2cos(x+y)}} \atop {cosxcosy=\frac{3}{4} }} \right.[/tex]

[tex]\left \{ {cosxcosy+sinxsiny=2cosxcosy-2sinxsiny}} \atop {cosxcosy=\frac{3}{4} }} \right.[/tex]

[tex]\left \{ {cosxcosy-3sinxsiny=0}} \atop {cosxcosy=\frac{3}{4} }} \right.[/tex]

[tex]\left \{ {\frac{3}{4} -3sinxsiny=0}} \atop {cosxcosy=\frac{3}{4} }} \right.[/tex]

[tex]\left \{ {sinxsiny=\frac{1}{4} }} \atop {cosxcosy=\frac{3}{4} }} \right.[/tex]

[tex]\left \{ {sinx^{2} siny^{2}=\frac{1}{16} }} \atop {cosxcosy=\frac{3}{4} }} \right.[/tex]

[tex]\left \{ {1-cosy^2-cosx^2+cosx^2cosy^2=\frac{1}{16} }} \atop {cosxcosy=\frac{3}{4} }} \right.[/tex]

[tex]\left \{ {cosy^2+cosx^2=\frac{3}{2} }} \atop {cosxcosy=\frac{3}{4} }} \right.[/tex]

[tex]\left \{ {cosy^2+(\frac{3}{4cosy})^2=\frac{3}{2} }} \atop {cosx=\frac{3}{4cosy} }} \right.[/tex]

[tex]\left \{ {cosy=+-\frac{\sqrt{3} }{2} }} \atop {cosx=\frac{3}{4cosy} }} \right.[/tex]

[tex]\left \{ {cosy=+-\frac{\sqrt{3} }{2} }} \atop {cosx=+-\frac{\sqrt{3} }{2} }} \right.[/tex]