1)tg^2x+ctg^2x+3tgx+3ctgx=-4 Решите плиз

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Nastyabuchukin

13 дек. 2013 г., 7:40:20 (10 лет назад)

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=0
tgx+ctgx=a
a²+3a+2=0
a1+a2=-3 U a1*a2=2
a1=-2⇒tgx+ctgx=-2
tgx+1/tgx+2=0
tg²x+2tgx+1=0  tgx≠0
(tgx+1)²=0⇒tgx=-1⇒x=-π/4+πn
a2=-1⇒tgx+ctgx=-1
tgx+1/tgx+1=0
tg²x+tgx+1=0  tgx≠0
tgx=t⇒t²+t+1=0
D=1-4=-3-решения нет
2)(1-сos2x)/2+(1-cos4x)/2=(1+cos6x)/2+(1+cos8x)/2
1-cos2x+1-cos4x=1+cos6x+1+cos8x
cos8x+cos6x+cos4x+cos2x=0
2cos5xcos3x+2cos5xcosx=0
2cos5x(cos3x+cosx)=0
2cos5x*2cos2xcosx=0
4cos5xcos2xcosx=0
cos5x=0⇒5x=π/2+πn⇒x=π/10+πn/5
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2
cosx=0⇒x=π/2+πn