Question

Найдите абсциссы точек пересечения графиков функций y= tg(3-4x) y=ctg(5-x)

Answer 1

$tg (\, 3-4x) = \frac{1}{tg (\, 5-x)} \\ \\ tg (\, 3-4x) \cdot tg (\, 5-x) = 1 \\ \\ \frac{\sin (3-4x)}{\cos (3-4x)} \cdot \frac{\sin (5-x)}{\cos(5-x)}-1=0 \\ \\ \frac{\sin (3-4x) \cdot \sin(5-x)-\cos(3-4x) \cdot \cos(5-x)}{\cos(3-4x) \cdot \cos(5-x)}=0; \\\\ \frac{-(\cos(3-4x) \cdot \cos(5-x)- \sin (3-4x) \cdot \sin(5-x))}{\cos(3-4x) \cdot \cos(5-x)}=0; \ \ \ \ \frac{-\cos(3-4x + 5 -x)}{\cos(3-4x) \cdot \cos(5-x)}=0\\ \\ -\cos(-5x +8)=0; \ \ \ \cos(-5x+8) = 0 \\ \\ -5x+8=\frac{\pi}{2} + \pi n, \ n \in Z$

$5x-8=-\frac{\pi}{2} - \pi n, \ n \in Z \\ \\ 5x=-\frac{\pi}{2}+8 - \pi n, \ n \in Z; \ \ \ 5x=\frac{16- \pi}{2} - \pi n, \ n \in Z \\ \\ \boxed{x=\frac{16 - \pi}{10} - \frac{\pi n}{5}, \ n \in Z}$