Question

Решите уравнение sin3x+cos3x=0. найдите корни из отрезка [-pi/2 ; pi ]

Answer 1

$\sin 3x+\cos 3x=0|:\cos 3x\\ \\ tg3x+1=0\\ \\ tg3x=-1\\ \\ 3x=- \frac{\pi}{4} + \pi n,n \in Z|:3\\ \\ x=- \frac{\pi}{12}+ \frac{\pi n}{3},n \in Z$

Отбор корней:

$n=0;\,\,\,\, x=-\frac{\pi}{12}\\ n=1;\,\,\,\, x=-\frac{\pi}{12}+ \frac{\pi}{3}=\frac{\pi}{4}\\ n=2;\,\,\,\, x=-\frac{\pi}{12}+\frac{2\pi}{3}=\frac{7\pi}{12}\\ n=3;\,\,\,\,x=-\frac{\pi}{12}+ \pi =\frac{11\pi}{12}\\ n=-1;\,\,\,\,x=-\frac{\pi}{12}-\frac{\pi}{3}=-\frac{5\pi}{12}$