Помогите решить алгебру

Ответ

Автор - saddatimov

282.3

$left { {{sin{x}sin{y}=frac{3}{4}} atop {tan{x}tan{y}=3}} right. \\tan{x}tan{y} = frac{sin{x}}{cos{x}} cdot frac{sin{y}}{cos{y}} = 3 Rightarrow cos{x}cos{y} = frac{1}{4} \left { {{sin{x}sin{y}=frac{3}{4}} atop {cos{x}cos{y} = frac{1}{4}} right. \\cos{x}cos{y} + sin{x}sin{y} = cos{(x-y)} = frac{3}{4}+frac{1}{4}=1 Rightarrow x-y=2pi n, n in mathbb{Z} \x = 2pi n+y \sin{(2pi n + y)sin{y}}=frac{3}{4} \sin^2{y} = frac{3}{4} \$ $y = pm frac{pi}{3}cdot (-1)^k + pi k \x = 2pi n + y = pm frac{pi}{3}cdot (-1)^k + pi k + 2pi n = pm frac{pi}{3}cdot (-1)^k + pi(k+2n) \k,n in mathbb{Z} \$

285.4

$left { {{x+y=frac{pi}{4}} atop {tan{x}tan{y}=frac{1}{6}}} right. \\x=frac{pi}{4}-y \tan{x}tan{y} = tan{(frac{pi}{4}-y)}tan{y} = frac{tan{frac{pi}{4}}-tan{y}}{1+tan{frac{pi}{4}}cdot tan{y}} cdot tan{y} = frac{1}{6} \tan{y} = t \frac{1-t}{1+t}cdot t = frac{1}{6} \frac{t-t^2}{1+t}-frac{1}{6}=0 \$ $frac{t-t^2-6-6t}{1+t}=0 \frac{-t^2-5t-6}{1+t}=0 \t_1 = -2 qquad t_2=-3 \\left [ {{y=arctan{(-2)}+pi k quad x =-arctan{(-2)}+frac{pi}{4}-pi k} atop {y=arctan{(-3)}+pi n quad x =-arctan{(-3)}+frac{pi}{4}-pi n }} right. quad n, k in mathbb{Z}$

Остальные варианты решаются аналогично.