Question

Найдите решение уравнения на указанном интервале:
sin5x-cos4x=0 360°


cos6x-sin3x=0 90°

Answer 1

$1)sin(5x)-cos(4x)=0<=>sin(5x)-sin(\frac{\pi}{2}-4x )=0<=>\\<=>cos(\frac{2x+\pi}{4} )sin(\frac{18x-\pi}{4} )=0\\cos(\frac{2x+\pi}{4} )=0=>x=\frac{\pi}{2}+2\pi k\\ sin(\frac{18x-\pi}{4} )=0<=>18x-\pi=4\pi k<=>x=\frac{\pi}{18}+\frac{2\pi k}{9}\\x=\frac{\pi}{18}+\frac{2\pi k}{9}$

$2)cos(6x)-sin(3x)=0<=>sin(\frac{\pi}{2}-6x )-sin(3x)=0<=>\\<=>cos(\frac{\pi-6x}{4} )sin(\frac{\pi-18x}{4} )=0\\cos(\frac{\pi-6x}{4} )=0=>x=\frac{\pi}{2}+\frac{2\pi k}{3}\\  sin(\frac{\pi-18x}{4} )=0=>x=\frac{\pi}{18}+\frac{2\pi k}{9}\\x=\frac{\pi}{18}+\frac{2\pi k}{9}$