Phương trình lượng giác

0

037675582

sin^8(x)+cos^8(x)=17/16cos^2(2x)
=>[sin^4(x)+cos^4(x)]^2-2sin^4(x).cos^4(x)=17\16cos^2(2x)
<=>[1-1\2sin^2(2x)]^2-1\8sin^4(2x)=17\16cos^2(2x)
<=>1-sin^2(2x)+1\8sin^4(2x)=17\16[1-sin^2(2x)]
<=>1\16sin^2(2x)+1\8sin^4(2x)-1\16=0
<=>sin^2(2x)=1\2 =>(1-cos4x)\2=1\2 <=>cos(4x)=0 <=>4x=pi\2+kpi =>x=pi\8+kpi\4
 
L

loigoimoi

cảm ơn ban nhiu nhé. minh hỏi ty nha (sin^4x+cox^4x)^2 dung cong thuc nao ra [1-1/2sin^2(2x)]^2 zay ban
 
You must log in or register to reply here.
Top Bottom