Question

помогите алгебра 10 класс 25 балов​

Answer 1

$\displaystyle \left \{ {{sin(x)*sin(y)=\frac{\sqrt{3}}{4}} \atop {cos(x)*cos(y)=\frac{\sqrt{3}}{4}}} \right.$

сложим уравнения

$\displaystyle cos(x)*cos(y)+sin(x)*sin(y)=\frac{\sqrt{3}}{2}\\\\cos(x-y)=\frac{\sqrt{3}}{2}\\\\x-y= \pm \frac{\pi}{6}+2\pi n; n \in Z\\\\x=y \pm \frac{\pi }{6}+2\pi n$

1) x=y+π/6

$\displaystyle sin(y+\pi /6)*sin(y)=\frac{\sqrt{3}}{4}\\\\\frac{1}{2}(cos (\pi /6)-cos(2y+\pi /6))=\frac{\sqrt{3}}{4}\\\\\frac{\sqrt{3}}{2}-cos(2y+\pi /6)=\frac{\sqrt{3}}{2}\\\\cos (2y+\pi /6)=0\\\\2y+\pi /6= \pm \frac{\pi}{2}+2\pi n$

1.1 ) 2y+π/6= π/2+2πn

$\displaystyle 2y=\pi /2-\pi /6+2\pi n=\pi /3+2\pi n\\\\y_1=\pi /6+\pi n; x_1=\pi /6+\pi /6+2\pi n=\pi /3+2\pi n$

1.2)  2y+π/6= -π/2+2πn

$\displaystyle 2y= -\pi /6-\pi /2+2\pi n=-2\pi/3+2\pi n\\\\\ y_2=-\pi /3+2\pi n ; x_2=-\pi /3+\pi /6+2\pi n=-\pi /3+2\pi n$

ответом на 1 случай (π/3+2πn; π/6+2πn) или (-π/3+2πn; -π/3+2πn)

2) x= y-π/6

решаем аналогично

$\displaystyle cos (2y-\pi /6)=0\\\\2y-\pi /6= \pm \pi /2+2\pi n$

2.1) 2y-π/6=π/2+2πn

$\displaystyle 2y=\pi /6+\pi /2+2\pi n=2\pi /3+2\pi n\\\\y_3=\pi /3+2\pi n: x_3=\pi /3-\pi /6+2\pi n=\pi /6+2\pi n$

2.2) 2y-π/6= -π/2+2πn

$\displaystyle 2y=\pi /6-\pi /2+2\pi n= -\pi /3+2\pi n\\\\\\y_4= -\pi /6+2\pi n; x_4=-\pi /6-\pi /6+2\pi n= -\pi /3+2\pi n$

ответом на 2 случай (π/6+2пn; π/3+2πn) или (-π/3+2πn; -π/6+2πn)