Автор - kzkzkz20030511

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


cos6x-sin3x=0 90°

Ответ

Автор - igundane

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

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