Giải các phương trình
LG a
cos2x+cos22x−cos23x−cos24x=0
Lời giải chi tiết:
cos2x+cos22x−cos23x−cos24x=0⇔1+cos2x2+1+cos4x2−1+cos6x2−1+cos8x2=0⇔1+cos2x+1+cos4x−1−cos6x−1−cos8x=0⇔(cos2x+cos4x)−(cos6x+cos8x)=0⇔2cos3xcosx−2cos7xcosx=0⇔2cosx(cos3x−cos7x)=0⇔2cosx.[−2sin5xsin(−2x)]=0⇔4cosxsin5xsin2x=0⇔[cosx=0sin5x=0sin2x=0⇔[x=π2+kπ5x=kπ2x=kπ⇔[x=π2+kπx=kπ5x=kπ2⇔[x=kπ5x=kπ2,k∈Z
LG b
cos4xcos(π+2x)−sin2xcos(π2−4x) =√22sin4x
Lời giải chi tiết:
cos4xcos(π+2x)−sin2xcos(π2−4x)=√22sin4x⇔−cos4xcos2x−sin2xsin4x=√22sin4x⇔−(cos4xcos2x+sin2xsin4x)=√22sin4x⇔−cos(4x−2x)=√22.2sin2xcos2x⇔−cos2x=√2sin2xcos2x⇔√2sin2xcos2x+cos2x=0⇔cos2x(√2sin2x+1)=0⇔[cos2x=0√2sin2x+1=0⇔[2x=π2+kπsin2x=−1√2⇔[x=π4+kπ22x=−π4+k2π2x=5π4+k2π⇔[x=π4+kπ2x=−π8+kπx=5π8+kπ,k∈Z
LG c
tan(1200+3x)−tan(1400−x)=2sin(800+2x)
Lời giải chi tiết:
ĐK:
{1200+3x≠900+k.18001400−x≠900+k.1800⇔{3x≠−300+k.1800x≠500−k.1800⇔{x≠−100+k.1800x≠500−k.1800
tan(1200+3x)−tan(1400−x)=2sin(800+2x)⇔tan[3(400+x)]−tan[1800−(400+x)]=2sin[2(400+x)]⇔tan[3(400+x)]−tan[−(400+x)]=2sin[2(400+x)]⇔tan[3(400+x)]+tan(400+x)=2sin[2(400+x)]
Đặt 400+x=y ta được:
tan3y+tany=2sin2y⇔sin3ycos3y+sinycosy=2sin2y⇔sin3ycosy+sinycos3ycos3ycosy=2sin2ycos3ycosycos3ycosy⇒sin3ycosy+sinycos3y=2sin2ycos3ycosy⇔sin4y−2sin2ycos3ycosy=0⇔2sin2ycos2y−2sin2ycos3ycosy=0⇔2sin2y(cos2y−cos3ycosy)=0⇔2sin2y[cos2y−12(cos4y+cos2y)]=0⇔2sin2y(12cos2y−12cos4y)=0⇔sin2y(cos2y−cos4y)=0⇔sin2y.[−2sin3ysin(−y)]=0⇔2sinysin2ysin3y=0⇔[siny=0sin2y=0sin3y=0⇔[y=k18002y=k18003y=k1800⇔[y=k.1800y=k.900y=k.600⇔[y=k.600y=900+k.1800
Suy ra
[x+400=k.600x+400=900+k.1800⇔[x=−400+k.600x=500+k1800(loai)⇔x=−400+k.600,k∈Z
Vậy pt có nghiệm x=−400+k.600,k∈Z.
LG d
tan2x2+sin2x2tanx2+cos2x2.cot2x2+cot2x2+sinx=4
Lời giải chi tiết:
x = (4k + 1) π/2;
x = (-1)(k+1)arcsin2/3 + kπ.
LG e
sin2t+2cos2t−1cost−cos3t+sin3t−sint=cost
Lời giải chi tiết:
sin2t+2cos2t−1cost−cos3t+sin3t−sint=cost⇔sin2t+cos2t−2sin2tsin(−t)+2cos2tsint=cost⇔sin2t+cos2t2sin2tsint+2cos2tsint=cost⇔sin2t+cos2t2sint(sin2t+cos2t)=cost
ĐK: {sint≠0sin2t+cos2t≠0(*)
⇔12sint=cost⇒1=2sintcost⇔1=sin2t⇔2t=π2+k2π⇔t=π4+kπ(TM(∗))
Vậy pt có nghiệm t=π4+kπ,k∈Z.
