Ответы и объяснения

  • Участник Знаний
2014-01-03T22:45:05+00:00
2sinxcosx-tgπ/10cos²x+tgπ/10sin²x-sin²x-cos²x=0
(tgπ/10-1)sin²x+2sinxcosx-(tgπ/10+1)cos²x=0
Поделим на cos²x≠0
(tgπ/10-1)tg²x+2tgx-(tgπ/10+1)=0
a=tgx
(tgπ/10-1)a²+2a-(tgπ/10+1)=0
D=4+4*(tgπ/10-1)*(tgπ/10+1)=4+4tg²π/10-4=4tg²π/10
√D=2tgπ/10
a1=(-2-2tgπ/10)/2(tgπ/10-1)=(tgπ/10+1)/(1-tgπ/10)⇒tgx=(tgπ/10+1)/(1-tgπ/10), x=arctg(tgπ/10+1)/(1-tgπ/10)+πn
a2=(-2+2tgπ/10)/2(tgπ/10-1)=(tgπ/10-1)/(tgπ/10-1)⇒tgx=1, x=π/4+πn