Question

Срочно!!!!! Нужна помощь!!!!! 1) 0,5(cos5x+cos7x)-cos^22x+sin^23x=0 (в степени только по одной 2-ке) 2) 2(cos4x-sinxcos3x)=sin4x+sin2x

Answer 1

1)0,5*2cos6xcosx-1/2(1+cos4x)+1/2(1-cos6x)=cos6xcosx-1/2-1/2cos4x-1/2cos6x=
=cos6xcosx-1/2(cos6x+cos4x)=cos6xcosx-1/2*2cos5xcosx=cos6xcosx-cos5xcosx=
=cosx(cos6x-cos5x)=cosx*(-2sin5,5xsin0,5x)=0
cosx=0⇒x=π/2+πn
sin5,5x=0⇒5,5x=πn⇒x=2πn/11
sin0,5x=0⇒0,5x=πn⇒2πn
2)2cos4x-2sinxcos3x-2sin3xcosx=0
2cos4x-2(sinxcos3x+sin3xcosx)=0
2cos4x-2sin4x=0
2cos²2x-2sin²2x-4sin2xcos2x=0  /-2cos²2x≠0
tg²2x+2tgx-1=0
tgx=a
a²+2a-1=0
D=4+4=8
a1=(-2-2√2)/2=-1-√2⇒tgx=-(1+√2)⇒x=-arctg(1+√2)+πn
a2=-1+√2⇒tgx=√2-1⇒x=arctg(√2-1)+πn