Question

Sin5x+sin7x+sin9x=0
с уравнением

Answer 1

$sin5x+sin7x+sin9x=0\\\\sin7x+2\, sin7x\cdot cos2x=0\\\\sin7x\cdot (1+2\, cos2x)=0\\\\a)\; \; sin7x=0\; ,\; \; 7x=\pi n\; ,\; \; x=\frac{\pi n}{7}\; ,\; n\in Z\\\\b)\; \; cos2x=-\frac{1}{2}\; ,\; \; 2x=\pm arccos(-\frac{1}{2})+2\pi n\; ,\; n\in Z\\\\2x=\pm (\pi -arccos\frac{1}{2})+2\pi k=\pm (\pi -\frac{\pi}{3})+2\pi k\; ,\; k\in Z\\\\2x=\pm \frac{2\pi }{3}+2\pi k\; ,\; k\in Z\\\\x=\pm \frac{\pi }{3}+\pi k\; ,\; k\in Z\\\\Otvet:\; \; x=\frac{\pi n}{7}\; ,\; \; x=\pm \frac{\pi}{3}+\pi k\; ,\; n,k\in Z\; .$

Answer 2

sin5x+sin7x+sin9x=0

sin7x+(sin5x+sin9x)=0

sin7x+2sin7xcos2x=0

sin7x*(1+2cos2x)=0

sin7x=0; 7х= πn; n∈Z; x= πn/7; n∈Z;

1+2cos2x=0

cos2x=-0.5

2х=±arccos(-0/5)+2πn;  n∈Z

2х=±2π/3+2πn;  n∈Z

х=±π/3+πn;  n∈Z