1)cos^2x-cos^2x+cos^2(3x)-cos^2(4x)=0 2)sin^2(3x)+sin^2(4x)+sin^2(6x)+sin^2(7x)=2 - вопрос №2080685122232402232426 от irinalera22 30.05.2021 06:35

Question

Алгебра: 1)cos^2x-cos^2x+cos^2(3x)-cos^2(4x)=0 2)sin^2(3x)+sin^2(4x)+sin^2(6x)+sin^2(7x)=2 !

irinalera22 · Answer

0+(1++cos6x)/2-(1+cos8x)/2=0-(cos8x-cos6x)=02sin7xsinx=0sin7x=07x=πk   sinx=0x=πk/7  x=πkx2)1-cos6x+1-cos8x+1-cos12x+1-cos14x=4-(cos6x+cos14x)-(cos8x+cos12x)=02cos10xcos4x-2cos10xcos2x=0cos10x(cos4x+cos2x)=0x=π/20+πk/10cos3xcosx=0x=π/6+πkx=π/2+πk...