4cos ³x+3 √2sin2x=8cosx sin8x-cos6x= √3(sin6x+cos8x
2 câu trả lời
a) cosx(4cos2x+6√2sinx)=8cosx
<-> cosx = 0 hoac 4cos2x+6√2sinx=8
TH1: x=π/2+kπ
TH2: 4(1−sin2x)+6√2sinx=8
2sin2x−3√2sinx+2=0 (ptrinh vo nghiem)
Vay x=π/2+kπ.
b) sin(8x)−√3cos(8x)=√3sin(6x)+cos(6x)
1/2sin(8x)−√3/2cos(8x)=√3/2sin(6x)+1/2cos(6x)
sin(8x−π/3)=sin(6x+π/6)
8x−π/3=6x+π/6+2kπ hoac 8x−π/3=π−6x−π/6+2kπ
Hay x=π/3+kπ hoac x=π/12+kπ/7.
