Question

Система tgx+tgy=1
x+y=п/4​

Answer 1

(см. объяснение)

Объяснение:

$tg(x)+tg(y)=1\\x+y=\dfrac{\pi}{4}\\\\\dfrac{sin(x+y)}{cos(x)cos(y)}=1\\x+y=\dfrac{\pi}{4}\\\\\dfrac{\dfrac{\sqrt{2}}{2}}{cos(x)cos(y)}=1\\x+y=\dfrac{\pi}{4}\\\\\\\dfrac{cos(x-y)}{2}+\dfrac{cos(x+y)}{2}=\dfrac{\sqrt{2}}{2}\\x+y=\dfrac{\pi}{4}\\\\\\\dfrac{cos(x-y)}{2}+\dfrac{cos(\dfrac{\pi}{4})}{2}=\dfrac{\sqrt{2}}{2}\\x+y=\dfrac{\pi}{4}\\\\\\\dfrac{cos(x-y)}{2}=\dfrac{\sqrt{2}}{4}\\x+y=\dfrac{\pi}{4}\\\\\\cos(x-y)=\dfrac{\sqrt{2}}{2}\\x+y=\dfrac{\pi}{4}$

Найдем x-y:

$\left[\begin{array}{c}x-y=\dfrac{\pi}{4}+2k\pi,\; k\in Z\\x-y=\dfrac{7\pi}{4}+2k\pi,\; k\in Z\end{array}\right \\x+y=\dfrac{\pi}{4}\\\\1)\\x-y=\dfrac{\pi}{4}+2k\pi,\; k\in Z\\x+y=\dfrac{\pi}{4}\\\\\\2x=\dfrac{\pi}{2}+2k\pi,\; k\in Z\\x=\dfrac{\pi}{4}+k\pi,\; k\in Z\\y=-k\pi,\; k\in Z\\\\2)\\x-y=\dfrac{7\pi}{4}+2k\pi,\; k\in Z\\x+y=\dfrac{\pi}{4}\\\\\\x=\pi+k\pi,\; k\in Z\\y=-\dfrac{3\pi}{4}-k\pi, k\in Z$