Question

Решите уравнение: sin7x+sin5x+2sin6x=0

Answer 1

$sin(7x)+sin(5x)+2*sin(6x)=0\\\\ sin(7x)+sin(6x)+sin(6x)+sin(5x)=0\\\\ 2*sin(\frac{7x+6x}{2})*cos(\frac{7x-6x}{2})+2*sin(\frac{6x+5x}{2})*cos(\frac{6x-5x}{2})=0\\\\ sin(\frac{13x}{2})*cos(\frac{x}{2})+sin(\frac{11x}{2})*cos(\frac{x}{2})=0\\\\ \ \ [sin(\frac{13x}{2})+sin(\frac{11x}{2})]*cos(\frac{x}{2})=0\\\\ 2*sin(\frac{\frac{13x}{2}+\frac{11x}{2}}{2})*cos(\frac{\frac{13x}{2}-\frac{11x}{2}}{2})*cos(\frac{x}{2})=0$

$2*sin(\frac{\frac{13x}{2}+\frac{11x}{2}}{2})*cos(\frac{\frac{13x}{2}-\frac{11x}{2}}{2})*cos(\frac{x}{2})=0\\\\ sin(6x)*cos(\frac{x}{2})*cos(\frac{x}{2})=0\\\\ sin(6x)=0\ \ or\ \ cos(\frac{x}{2})=0\\\\ 6x=\pi k,\ k\in Z\ \ or\ \ \frac{x}{2}=\frac{\pi}{2}+\pi n,\ n\in Z\\\\ x=\frac{\pi k}{6},\ k\in Z\ \ or\ \ x=\pi+2\pi n,\ n\in Z\\\\ x=\frac{\pi k}{6},\ k\in Z$

Ответ: $\frac{\pi k}{6},\ k\in Z$