解二倍角方程

解方程4SinxCosx=3tan2x－6Tan2xSin^2 x，0≦x≦2π
Sol
4SinxCosx=3tan2x－6Tan2xSin^2x
Cos2x(4SinxCoSx)=cos2x(3Tan2x－6Tan2xSin^2x)
4Cos2xSinxcosx=3Sin2x－6Sin2xSin^2x
2Cos2xSin2x=3Sin2x－6Sin2xSin^2x
2Cos2xSin2x-3Sin2x+6Sin2xSin^2 x=0
Sin2x(2cos2x－3+6Sin^2 x)=0
Sin2x(2－3+4Sin^2 x)=0
Sin2x(4Sin^2 x－1)=0
Sin2x(2Sinx－1)(2Sinx+1)=0
Sin2x=0 or Sinx=1/2 or Sinx=－1/2
(1) Sin2x=0
0<=x<=2π
0<=2x<=4π
2x=0 orπor 2π or 3π or 4π
x=0 orπ/2 or π or 3π/2 or 2π
(2) Sinx=1/2
x=π/6 or 5π/6
(3) Sinx=－1/2
x=7π/6 or 11π/6

4sinxcox=3tan2x - 6tan2xsin^2x,0≦x≦2π
2sin(2x)=[3tan(2x)][1-2sin^2(x)]
2sin(2x)=[3tan(2x)]cos(2x)
2tan(2x)=3tan(2x)
tan(2x)=0
2x=0 or π or 2π or 3π or 4π
x=0 or π/2 or π or 3π/2 or 2π

2014-01-14 23:15:42 補充：
in line 6,
Sin2x(2cos2x－3+6Sin^2 x)=0
sin2x[2(1-2sin^2 x)-3+6sin^2 x]=0
sin2x(2-4sin^2 x-3+6sin^2 x)=0
sin2x(2sin^2 x-1)=0
-sin2xcos2x=0
sin4x=0
x=0 , π/4(reject) , π/2 , 3π/4(reject) , π , 5π/4(reject) , 3π/2 , 7π/4(reject) , 2π

general：nπ/2 , where n is a positive integer.