Question

Помогите решить алгебру

Answer 1

282.3

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

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.