1)tg^2x+ctg^2x+3tgx+3ctgx=-4 решите 2)sin^2x+sin^2(2x)=cos^2(3x)+cos^(4x) - вопрос №12233422222343131348 от 45667889 18.12.2021 10:26

Question

Алгебра: 1)tg^2x+ctg^2x+3tgx+3ctgx=-4 решите 2)sin^2x+sin^2(2x)=cos^2(3x)+cos^(4x)

45667889 · Answer

1)tg²x+ctg²x+3tgx+3cygx=-4(tgx+ctgx)²-2tgxctgx+3(tgx+ctgx)+4=0      tgx*ctgx=1(tgx+ctgx)²+3(tgx+ctgx)+2=0tgx+ctgx=aa²+3a+2=0a1+a2=-3 U a1*a2=2a1=-2⇒tgx+ctgx=-2tgx+1/tgx+2=0tg²x+2tgx+1=0  tgx≠0(tgx+1)²=0⇒tgx=-1⇒x=-π/4+πna2=-1⇒tgx+ctgx=-1tgx+1/tgx+1=0tg²x+tgx+1=0  tgx≠0tgx=t⇒t²+t+1=0D=1-4=-3-решения нет2)(1-сos2x)/2+(1-cos4x)/2=(1+cos6x)/2+(1+cos8x)/21-cos2x+1-cos4x=1+cos6x+1+cos8xcos8x+cos6x+cos4x+cos2x=02cos5xcos3x+2cos5xcosx=02cos5x(cos3x+cosx)=02cos5x*2cos2xcosx=04cos5xcos2xcosx=0cos5x=0⇒5x=π/2+πn⇒x=π/10+πn/5cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2cosx=0⇒x=π/2+πn...